UC-NRLF 


ELEMENTS  OF 

DESCRIPTIVE 
GEOMETRY 


$B    35    STO 


RANDALL 


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in  2007  with  funding  from 

Microsoft  Corporation 


http://www.archive.org/details/elennentsofdescriOOrandrich 


ELEMENTS  OF  DESCRIPTIVE 
GEOMETRY 


WITH  APPLICATIONS  TO 


Isometric  Projection  and  Other  Forms  of 
One-Plane  Projection 


A    TEXT-BOOK   FOR    COLLEGES  AND 
ENGINEERING    SCHOOLS 


BY 


0.  E.  RANDALL,  Ph.D. 

Pkofessor  of  Mechanical  Drawing,  Brown  University 


GINN  &  COMPANY 

BOSTON  •  NEW  YORK  •  CHICAGO  •  LONDON 


Copyright,  1905 
By  O.  E.  RANDALL 


ALL  RIGHTS   RESERVED 
55.9 


HCht  atftengum  ^regg 

GINN   &   COMPANY  •   PRO- 
PRIETORS •  BOSTON  .  U.S.A. 


-^3 


PREFACE 

The  aim  of  this  treatise  is  to  make  a  clear  presentation  of  the 
theory  of  projection,  to  show  the  application  of  this  theory  as  a 
medium  of  expression,  and  by  the  discussion  and  proof  of  a  great 
variety  of  problems  to  enable  the  student  to  make  a  ready  and 
intelligent  use  of  this  medium  in  the  representation  of  all  forms 
of  magnitudes. 

As  by  far  the  greater  part  of  practical  drafting  is  done  from 
the  standpoint  of  the  third  quadrant,  there  seems  to  be  no  good 
reason  why  the  principles  of  descriptive  geometry,  which  are  so 
directly  and  extensively  applied  in  practice,  should  not  also  be 
presented  from  the  standpoint  of  the  same  quadrant. 

Therefore,  while  the  student  is  called  upon  to  work  freely  in 
all  the  four  quadrants,  the  subject-matter  is  presented  primarily 
from  the  third  quadrant. 

In  the  establishment  of  principles  great  effort  is  made  to  be 
explicit ;  but  in  the  application  of  these  principles,  for  which 
purpose  a  great  many  unsolved  problems  are  assigned,  the  student 
is  left  largely  to  his  own  resources^ 

As  the  principles  of  projection  are  fundamental  in  all  branches 
of  drafting,  it  follows  that  no  attempt  at  extensive  application  of 
these  principles  in  such  subjects  as  machine  drawing,  gearing, 
architectural  drawing,  etc.,  should  be  made  until  the  principles 
themselves  have  been  thoroughly  established.  For  this  reason 
the  attention  of  this  work  is  largely  confined  to  theoretical  con- 
siderations, although  a  number  of  simple  practical  applications  such 

as  the  student  can  safely  and  intelligently  make  are  introduced. 

iii 


1^319100 


iv  PREFACE 

Free  use  is  made  of  profile  and  other  supplementary  planes  of 
projection. 

Isometric  projection  and  other  forms  of  one-plane  projection  are 
treated  as  applications  of  descriptive  geometry. 

It  is  hoped  that  the  system  of  notation  which  is  introduced 
will  be  found  both  simple  and  expressive;  and  that  the  method 
of  locating  given  parts  which  may  be  employed  in  the  assign- 
ment of  work  in  the  recitation  room  and  in  the  drafting  room 
will  be  found  usefiil. 


CONTENTS 


CHAPTER  PAGE 

1 


I.  Definitions  and  Assumptions    .         .         .         .     '   . 

II.  Representation  of  the  Point,  Line,  and  Plane  . 

III.  Supplementary  Planes  of  Projection     . 

IV.  Notation       .         .         . 

V.  Method  of  Locating  Given  Parts  .... 

VI.  Problems  relating  to  the  Point,  Line,  and  Plane 

VII.  Generation  and  Classification  of  Lines 

VIII.  Generation  and  Classification  of  Surfaces 

IX.  Representation  of  Surfaces  with  Plane  Faces  . 

X.  Representation  of  Single  Curved  Surfaces 

XL  Representation  of  Warped  Surfaces 

XII.  Representation  of  Surfaces  of  Revolution 

XIII.  Determination   of    Planes    Tangent    to    Surfaces    of 
Single  Curvature 


XIV.     Determination    of    Planes  Tangent    to 
Double  Curvature 


XV.     Intersection  of  Surfaces  by  Lines 

XVI.     Intersection  of  Surfaces  by  Planes 

XVII.     Intersection  of  Surfaces  by  Surfaces 

XVIII.     Isometric  Projection  and  Other  Forms 
Projection 


Surfaces   of 


OF  One-Plane 


5 
24 
32 
34 
36 
74 
84 
89 
93 
103 
115 

122 

144 
153 

158 
177 

193 


DESCRIPTIVE  GEOMETRY 

CHAPTER  I 

DEFINITIONS  AND   ASSUMPTIONS 

1.  The  Subject  defined.  Descriptive  geometry  is  that  branch  of 
mathematics  which  seeks,  through  the  medium  of  an  exact  pro- 
cess of  graphic  expression,  to  represent  geometrical  magnitudes 
which  occupy  given  positions  in  space,  and  also  through  the 
same  medium  of  expression  to  solve  such  problems  as  relate  to 
these  magnitudes. 

2.  Representation  of  Magnitudes  of  Two  Dimensions.  A  magni- 
tude of  two  dimensions,  such  as  a  plane  geometrical  figure,  may  be 
easily  and  directly  represented,  graphically,  upon  a  single  plane, 
since  every  characteristic  of  such  a  magnitude  may  be  determined 
from  a  single  standpoint  of  observation,  and  the  whole  may  be 
outlined  upon  the  very  plane  in  which  the  magnitude  exists. 

The  diagrams  connected  with  the  statement  and  solution  of 
problems  in  plane  geometry  furnish  an  illustration  of  this  fact. 

3.  Representation  of  Magnitudes  of  Three  Dimensions.  A  magni- 
tude of  three  dimensions  does  not  exist  in  a  single  plane,  neither 
can  its  characteristics  be  completely  determined  from  a  single 
standpoint  of  observation ;  therefore  the  process  of  representation 
must  necessarily  be  different  from  that  employed  in  connection 
with  magnitudes  of  two  dimensions. 

4.  Projection.  Since  the  points  and  lines  of  magnitudes  of  three 
dimensions  do  not  exist  in  a  single  plane,  as  is  the  case  with 
magnitudes  of  two  dimensions,  it  will  be  necessary  to  determine 
some  plane  of  representation  and  to  establish  some  process  by 
which  reference  to  this  plane  may  be  made. 

1 


2  DESCRIPTIVE  GEOMETRY 

The  plane  of  representation,  or  the  plane  upon  which  the  repre- 
sentation is  made,  is  called  the  plane  of  projection^*  and  the  process 
by  which  reference  to  this  plane  is  made  is  called  projection. 

Projection  involves  the  assumption  of  a  point  of  sight,  or  posi- 
tion for  the  observer,  and  a  plane  of  projection. 

When  the  point  of  sight  and  the  plane  of  projection  are  given, 
the  projection  of  an  object,  for  simplicity  a  point,  is  accomplished 
when  the  point  is  literally  thrown  forward  along  its  visual  rayf 
until  it  rests  upon  the  plane.  In  other  words,  the  projection  of  a 
point  upon  any  plane  is  the  intersection  of  the  visual  ray  of  the 
point  with  that  plane. 

The  point  may  occupy  a  position  between  the  observer  and  the 
plane  of  projection,  or  the  plane  of  projection  may  stand  between 
the  observer  and  the  point;  but  in  either  case  the  projection  of 
the  point  is  found  by  the  process  stated  above. 

If  the  point  is  in  the  plane  of  projection,  it  is  evident  that  the 
point  and  its  projection  will  be  identical. 

5.  Systems  of  Projection.  It  is  evident  that  the  character  of  the 
projection  of  a  magnitude,  which  consists  of  a  collection  of  points, 
will  depend  upon  the  relative  positions  of  the  magnitude,  the 
observer,  and  the  plane  of  projection. 

There  are  two  principal  systems  of  projection,  depending  upon 
the  position  of  the  observer  with  reference  to  the  plane  of  projec- 
tion, —  the  scenographic  projection  and  the  orthographic  projection. 

The  scenographic  projection  is  that  system  in  which  the  point 
of  sight  is  assumed  within  a  finite  distance  of  the  plane  of 
projection. 

As  this  is  the  position  which  would  naturally  be  assumed  by  an 
observer,  the  projection  upon  the  plane  will  correspond  with  that 
made  upon  the  retina  of  the  eye,  and  the  picture  will  be  true  to 
nature.  This  system  is  employed  whenever  it  is  desired  to  repre- 
sent an  object  as  it  appears,  rather  than  to  show  its  exact  dimen- 
sions ;  but  on  account  of  difficulties  attending  the  operation  of  the 

*  Projection  may  be  made  upon  various  surfaces,  such  as  cylindrical  surfaces, 
spherical  surfaces,  etc.  ;  but  in  this  work  attention  will  be  confined  to  projection 
upon  plane  surfaces. 

t  A  visual  ray  is  any  straight  line  passing  through  the  point  of  sight. 


DEFINITIONS  AND  ASSUMPTIONS  3 

system,  owing  to  the  obliquity  of  the  visual  rays  to  the  plane  of 
projection,  the  system  is  not  practicable  for  problematic  work. 

The  orthographic  projection — the  system  which  will  be  employed 
throughout  this  work  —  is  that  system  in  which  the  point  of  sight 
is  assumed  at  an  infinite  distance  from  the  plane  of  projection. 

In  this  system,  since  magnitudes  are  assumed  within  a  finite 
distance  of  the  plane  of  projection,  visual  rays  to  the  various 
points  of  such  magnitudes  may  be  regarded  as  parallel  lines,  and 
may  be  assumed  perpendicular  to  the  plane  of  projection. 

Under  these  conditions  the  orthographic  projection  of  a  point 
upon  any  plane  is  the  point  in  which  a  straight  line  drawn  through 
the  point  perpendicular  to  the  plane  pierces  the  plane. 

6.  Planes  of  Projection.  As  a  rule  it  is  not  possible  to  learn  all 
the  characteristics  of  magnitudes  of  three  dimensions  from  a  single 
standpoint  of  observation,  and  for  this  reason  more  than  one  plane 
of  projection  is  usually  needed. 

There  are  two  principal  planes  of  projection,  —  a  horizontal  plane 
called  the  horizontal  plane  of  projection  or  H,  and  a  vertical  plane 
perpendicular  to  H  and  called  the  vertical  plane  of  projection  or  V. 

7.  The  Ground  Line.  The  intersection  of  H  and  V  is  called  the 
ground  line,  or  G-L. 

8.  Quadrants.  The  planes  H  and  V  divide  space  into  four  right 
dihedral  angles  know^n  as  the  first  quadrant,  the  second  quadrant, 
the  third  quadrant,  and  the  fourth  quadrant. 

The  first  quadrant  is  above  H  and  in  front  of  V,  the  second 
quadrant  is  above  H  and  back  of  V,  the  third  quadrant  is  below  II 
and  back  of  V,  and  the  fourth  quadrant  is  below  II  and  in  front  of  V. 

9.  Projecting  Lines.  In  orthographic  projection  the  visual  rays 
are  called  projecting  lines,  and  are  assumed  perpendicular  to  the 
plane  on  which  projection  is  made. 

10.  Position  of  the  Observer.  When  projecting  on  ff  the  observer 
is  supposed  to  be  above  //  and  at  an  infinite  distance  from  it. 

When  projecting  on  V  the  observer  is  supposed  to  be  in  front 
of  V  and  at  an  infinite  distance  from  it. 

11.  Horizontal  and  Vertical  Projections.  Projections  on  II  are 
called  horizontal  projections,  and  projections  on  V  are  called  vertical 
projections. 


4  DESCEIPTIVE  GEOMETRY 

12.  Revolution  of  the  Planes  of  Projection.  The  primary  posi- 
tion of  H  is  horizontal,  and  the  primary  position  of  V  is  vertical, 
giving  horizontal  and  vertical  projections  on  two  distinct  planes 
perpendicular  to  each  other. 

In  order  that  projections  upon  //  and  V  may  be  represented  upon 
a  single  plane,  one  of  the  planes  of  projection  is  revolved  about 
G-L  as  an  axis  until  it  is  coincident  with  the  other. 

This  last  position  of  H  and  V  is  called  the  secondary  position, 
or  position  of  coincidence. 


CHAPTER  II 


REPRESENTATION  OF  THE  POINT,  LINE,  AND  PLANE 

13.  Projection  of  the  Point  upon  H  and  V  in  their  Primary  Position. 

In  Fig.  1  let  H  and  V  represent  the  horizontal  and  vertical  planes 
of  projection  in  their  primary  position,  and  let  M  represent  a  point 
in  the  first  quadrant. 
Then,  according  to  Sec- 
tion 5,  the  horizontal 
projection  of  Jf,  or  the 
projection  of  M  upon  iT, 
is  iTij, ;  and  the  vertical 
projection  of  i¥,  or  the 
projection  of  M  upon  F, 
is  m".  The  line  M-m,,  is 
perpendicular  to  H  and 
is  called  the  horizontal 
projecting  line  of  M. 
The  line  M-m"  is  perpen- 
dicular to  V  and  is  called  the  vertical  projecting  line  of  M, 

iV  is  a  point  in  the  second  quadrant.  Its  horizontal  projection 
is  n,,  and  its  vertical  projection  is  n".  The  line  N-n^f  is  perpen- 
dicular to  H  and  is  called  the  horizontal  projecting  line  of  N,  The 
line  N-n''  is  perpendicular  to  V  and  is  called  the  vertical  project- 
ing line  of  iV. 

O  is  a  point  in  the  third  quadrant.  Its  horizontal  projection  is 
o^p  and  its  vertical  projection  is  o".  The  line  O-o^^  is  perpendic- 
ular to  H  and  is  called  the  horizontal  'projecting  line  of  0.  The 
line  0-(?"  is  perpendicular  to  V  and  is  called  the  vertical  projecting 
line  of  0. 

P  is  a  point  in  the  fourth  quadrant.  Its  horizontal  projection  is 
jt?„  and  its  vertical  projection  is  p".  The  line  P-p,,  is  perpendic- 
ular to  H  and  is  called  the  liorizontal  projecting  line  of  P.    The 


Fig.  1 


6  DESCRIPTIVE  GEOMETRY 

line  P-p"  is  perpendicular  to  V  and  is  called  the  vertical  projecting 
line  of  F. 

It  will  be  noticed  in  Fig.  1  that  the  horizontal  and  vertical  pro- 
jecting lines  of  a  point  determine  a  plane  which  is  perpendicular 
to  both  H  and  V  and  is  therefore  perpendicular  to  G-L ;  also  that 
the  straight  lines  in  which  this  plane  intersects  H  and  V  are  per- 
pendicular to  G-L  at  the  same  point  and  pass  respectively  through 
the  horizontal  and  vertical  projections  of  the  point. 

Observe  that  when  the  horizontal  and  vertical  projections  of  a 
point  are  given,  the  point  itself  is  definitely  located,  for  the  hori- 
zontal and  vertical  projecting  lines,  determined  by  the  projections  of 
the  point,  lie  in  the  same  plane  and  intersect  at  the  only  point  which 
can  have  its  horizontal  and  vertical  projections  at  the  points  given. 

Observe  that  the  distance  of  a  point  from  H  is  in  each  case 
indicated  by  the  distance  of  its  vertical  projection  from  G^-i,  and 
that  the  distance  of  the  point  from  V  is  in  each  case  indicated  by 
the  distance  of  its  horizontal  projection  from  G-L. 

If  a  point  is  situated  in  H^  its  horizontal  projection  is  the  point 
itself  and  its  vertical  projection  is  in  G-L, 

If  a  point  is  situated  in  F,  its  vertical  projection  is  the  point 
itself  and  its  horizontal  projection  is  in  G-L. 

If  a  point  is  situated  in  G-L^  both  its  horizontal  and  vertical 
projections  coincide  with  the  point  itself. 

14.  Representation  of  the  Point  upon  H  and  V  in  their  Position  of 
Coincidence.  In  Fig.  2  the  projections  m,,,  m'^;  riff^n";  Of,,o";  and 
Pfn  P"  ^^®  those  previously  found  in  Fig.  1  and  represent  points  in 
the  first,  second,  third,  and  fourth  quadrants  respectively. 

If  the  plane  H.  be  revolved  about  G-L  as  an  axis  until  that 
portion  of  H  back  of  V  falls  on  V  above  //,  and  that  portion  of  If 
in  front  of  V  falls  on  V  below  ff,  the  point  m"  will  remain  station- 
ary, while  the  point  m,,  will  move  in  the  arc  of  a  circle  with  a  as 
a  center,  and  fall  at  m,  in  the  line  m"-a  produced. 

The  point  n"  will  remain  stationary,  while  the  point  n,,  will 
move  in  the  arc  of  a  circle  with  5  as  a  center,  and  fall  at  n^  in  the 
line  6-n"  produced.  The  point  o"  will  remain  stationary,  while 
the  point  o,,  will  move  in  the  arc  of  a  circle  with  (^  as  a  center, 
and  fall  at  o^  in  the  line  o"-d  produced. 


THE  POINT,   LINE,   AND  PLANE 


•     y'""  y 

t 

jm  J     ^ 

L^;^^^'^ 

in-,t/^ 

\^.^^o„ 

/^  / 

Fig.  2 


The  point  ^''  will  remain  stationary,  while  the  point  jo,,  will 
move  in  the  arc  of  a  circle  with  e  as  a  center,  and  fall  at  jt?,  in  the 
line  e-p''. 

It  will  be  noticed  that  in  the  revolution  vertical  projections 
remain  stationary,  while  horizontal  projections  move  in  arcs  of 
circles  with  centers  in 
G-L^  and  fall  upon  V  in 
straight  lines  drawn 
through  the  correspond- 
ing vertical  projections 
perpendicular  to  G-L, 

After  the  revolution  of 
the  plane  H^  or  when  the 
planes  are  in  their  posi- 
tion of  coincidence,  the 
projections  will  appear  as 
shown  in  Fig.  3,  where 
that  portion  of  the  plane 
of  the  paper  above  G-L^ 
regarding  the  plane  of  the  paper  as  vertical,  represents  both  that 
portion  of  V  above  H  and  that  portion  of  H  back  of  F,  and  where 
that  portion  of  the  plane  of  the  paper  below  G-L  represents  both 
yO,  that  portion  of    V  below 

H  and  that  portion  of  H 
in  front  of  V. 

Again,  referring  to 
Fig.  2,  if  the  plane  V  be 
revolved  about  G-L  as  an 
axis  until  that  portion  of 
V  above  //  falls  upon  that 
portion  of  iiTback  of  F,  and 
yj(j  3  that  portion  of  V  below  // 

falls  upon  that  portion  of 
H  in  front  of  F,  the  horizontal  projections  will  remain  stationary, 
while  the  vertical  projections  will  move  in  arcs  of  circles  with 
centers  in  G-L^  and  fall  upon  H  in  straight  lines  drawn  through 
the  corresponding  horizontal  projections  perpendicular  to  G-L, 


1 
j 

V above  H     \ 

n, 

rm' 

H  behind  V  \ 

r                 ' 

io' 

P' 

V  below  H 

H  in  front  of 

^rrif 


8 


DESCRIPTIVE   GEOMETRY 


rUf 


m, 


Fio.  4 


After  the  revolution  of  the  plane  V  into  coincidence  with  // 
the  projections  will  again  be  expressed  as  shown  in  Fig.  3,  where 
that  portion  of  the  plane  of  the  paper  back  of  G-L^  regarding  the 
plane  of  the  paper  as  horizontal,  represents  both  that  portion  of 
M  behind  V  and  that  portion  of  V  above  H,  and  where  that  portion 
of  the  plane  of  the  paper  in  front  of  G-L  represents  both  that  por- 
tion  of   If  in   front   of  V  and   that 

tO. 

portion  of  V  below  //. 

It  will  be  observed  that  whether 
we  revolve  H  into  coincidence  with 
V  or  whether  we  revolve  V  into  coin- 
cidence with  H  the  result  will  be 
the  same.  When  working  on  the 
blackboard,  where  the  surface  is 
usually  vertical,  the  former  method 
will  be  found  convenient.  When 
working  on  the  drawing  board,  where 
the  surface  is  usually  horizontal,  the  latter  method  will  be  found 
more  natural. 

Again  referring  to  Fig.  1,  let  us  first  project  the  four  points 
Jf,  iV,  0,  and  P  upon  H  alone.  As  the  plane  V  is  perpendicular 
to  H  it  will  in  this  projection  appear  as  a  straight  line  coincident 
with  G-L.,  and  the  four  projections  will  appear  as  shown  in  Fig.  4, 
where  that  portion  of  the  plane  of  ^^, 

the  paper  back  of  G-L  represents  IT 
back  of  F,  and  where  that  portion  of 
the  plane  of  the  paper  in  front  of  G—L    ^ — 
represents  H  in  front  of  V.  i^, 

Now  project  the  four  points  upon 
V  alone.  The  plane  H  in  this  projec- 
tion will  appear  as'  a  straight  line 
coincident  with  G-L,  and  the  four  pi'ojections  will  appear  as  shown 
in  Fig.  5,  where  that  portion  of  the  plane  of  the  paper  above  G-L 
represents  V  above  //,  and  where  that  portion  of  the  plane  of  the 
paper  below  G-I^  represents  V  below  H. 

Now  if  we  place  Fig.  5  upon  Fig.  4  in  such  a  way  that  the 
ground  line  of  one  shall  coincide  with  the  ground  line  of  the 


P' 

Fig.  5 


THE  POINT,   LINE,   AND  PLANE  9 

other,  and  so  that  the  two  projections  of  each  point  shall  fall  upon 
the  same  straight  line  perpendicular  to  G-L^  the  result  will  be 
precisely  the  same  as  that  expressed  in  Fig.  3. 

It  will  be  observed  that  whether  we  revolve  H  into  coincidence 
with  F,  or  V  into  coincidence  with  H^  or  whether  the  projections 
are  made  upon  H  and  V  independently  and  afterwards  combined 
as  shown  above,  the  result  is  the  same. 

The  object  of  the  transformation  is  to  make  it  possible  to  repre- 
sent upon  a  single  plane,  projections  which  belong  primarily  upon 
two  planes  perpendicular  to  each  other. 

It  is  immaterial  by  which  method  the  transformation  is  made ; 
the  essential  thing  is  that  the  student  shall  be  able  to  pass  in  imagi- 
nation, without  any  difficulty  or  hesitation,  from  the  position  of 
perpendicularity  to  that  of  coincidence,  and  vice  versa. 

Returning  now  to  Fig.  3,  which  expresses  the  common  result  of 
the  three  methods  of  transformation,  it  will  be  noticed  that  when 
a  point,  as  Jf,  is  in  the  first  quadrant,  its  horizontal  projection  will 
be  in  front  of  G-L  and  its  vertical  projection  will  be  above  G-L ; 
that  when  a  point,  as  iV,  is  in  the  second  quadrant,  its  horizontal 
projection  will  be  back  of  G-L  and  its  vertical  projection  will  be 
above  G-L ;  that  when  a  point,  as  0,  is  in  the  third  quadrant,  its 
horizontal  projection  will  be  back  of  G-L  and  its  vertical  projec- 
tion will  be  below  G-L ;  that  when  a  point,  as  P,  is  in  the  fourth 
quadrant,  its  horizontal  projection  will  be  in  front  of  G-L  and  its 
vertical  projection  will  be  below  G-L. 

Conversely,  referring  to  Fig.  3,  if  a  horizontal  projection,  as  w,, 
is  situated  in  front  of  G-L.,  it  will  be  known  that  the  point  M 
must  be  in  front  of  F,  that  is,  either  in  the  first  quadrant  or  in  the 
fourth  quadrant ;  and  if  the  vertical  projection,  as  m',  of  the  same 
point  M  is  situated  above  (?-Z,  it  will  be  known  that  the  point  M 
must  be  above  H^  that  is,  either  in  the  first  quadrant  or  in  the 
second  quadrant,  and  therefore  it  will  be  known  that  the  point  M 
must  be  in  the  first  quadrant. 

If  a  horizontal  projection,  as  m,,  is  situated  back  of  G-L^  it  will 
be  known  that  the  point  N  must  be  back  of  F,  that  is,  either  in 
the  second  quadrant  or  in  the  third  quadrant;  and  if  the  corre- 
sponding vertical  projection,  as  n\  is  situated  above  G-L^  it  will  be 


10  DESCEIPTIVE  GEOMETRY 

known  that  the  point  JV  must  be  above  JI,  that  is,  either  in  the 
first  quadrant  or  in  the  second  quadrant,  and  therefore  it  will  be 
known  that  the  point  JV  must  be  in  the  second  quadrant. 

If  a  horizontal  projection,  as  o,,  is  situated  back  of  G—L,  it  will 
be  known  that  the  point  0  must  be  back  of  F,  that  is,  either  in 
the  second  quadrant  or  in  the  third  quadrant;  and  if  the  corre- 
sponding vertical  projection,  as  o',  is  situated  below  G-L,  it  will  be 
known  that  the  point  0  must  be  below  JJ,  that  is,  either  in  the 
third  quadrant  or  in  the  fourth  quadrant,  and  therefore  it  will  be 
known  that  the  point  0  must  be  in  the  third  quadrant. 

If  a  horizontal  projection,  as  jt?,,  is  situated  in  front  of  G-L,  it 
will  be  known  that  the  point  F  must  be  in  front  of  V,  that  is, 
either  in  the  first  quadrant  or  in  tlie  fourth  quadrant;  and  if  the 
corresponding  vertical  projection,  as  />',  is  situated  below  G-L^  it 
will  be  known  that  the  point  P  must  be  below  i/,  that  is,  either  in 
the  third  quadrant  or  in  the  fourth  quadrant,  and  therefore  it  will 
be  known  that  the  point  F  must  be  in  the  fourth  quadrant. 

It  will  be  observed,  from  a  comparison  of  Figs.  1,  2,  and  3,  that 
the  horizontal  and  vertical  projections  of  a  point,  in  the  trans- 
formed position  shown  in  Fig.  3,  must  lie  in  the  same  straight  line 
perpendicular  to  G-L. 

It  will  be  also  observed  that  in  the  transformed  position  the 
distance  of  the  horizontal  projection  of  a  point  from  G-L  still 
indicates  the  distance  of  the  point  itself  from  F,  and  that  the 
distance  of  the  vertical  projection  of  -a  point  from  G-L  still 
indicates  tlie  distance  of  the  point  itself  from  //. 

15.  To  assume  a  Point  at  Random.  To  assume  a  point  at  random, 
assume  its  projections  at  random,  provided  they  fall  in  the  same 
straight  line  perpendicular  to  G-I^. 

16.  Problem  1.  Determine  the  projections  of  a  point  situated  in 
the  first  quadrant,  2  units  from  H  and  4  units  from  V. 

17.  Problem  2.  Determine  the  projections  of  a  point  in  the  third 
quadrant,  1  unit  from  H  and  ^  units  from  V. 

18.  Problem  3.  Determine  the  projections  of  a  point  situated  in  H 
and  3  units  hack  of  V. 

19.  Problem  4.  Determine  the  projections  of  a  point  situated  in 
V  and  5  units  heloiv  If. 


THE  POINT,   LINE,   AND  PLANE 


11 


20.  Problem  5.  Determine  the  projections  of  a  point  situated  in 
the  fourth  quadrant,  4  units  from  If  and  4  units  from  V. 

21.  Problem  6.  —  Determine  the  projections  of  a  point  situated  in 
the  second  quadrant,  4  units  from  H  and  2  units  from  V, 

22.  Problem  7.  Determine  the  projections  of  a  point  situated  in 
G-L. 

23.  Projection  of  the  Straight  Line  upon  H  and  V  in  their 
Primary  Position.  As  a  straight  line  may  be  regarded  as  made  up 
of  an  infinite  number  of  consecutive  points,  the  projection  of  the 
straight  line  will  be  the  locus  of  the  projections  of  these  points. 

The  projecting  lines  of  these  points,  since  they  are  drawn  from 
points  in  a  straight  line  and  perpendicular  to  a  plane,  constitute 
a  plane  perpendicular  to  the  plane  of  projection. 

The  plane  made  up  of  these  projecting  lines  is  called  the  pro- 
jecting plane  of  the  line.  If  the  projection  is  being  made  upon 
H,  the  projecting  plane  is  called  the  horizontal  projecting  plane; 
and  if  the  projection  is  being  made  upon  F,  the  projecting  plane 
is  called  the  vertical  2yrojecting  plane. 

The  projection  of  a  straight  line  upon  a  plane,  then,  is  a  straight 
line,  and  for  this  reason  its  position  is  fully  known  when  two  of 
its  points  are  determined. 

In  Fig.  6,  which  shows 
//  and  V  in  their  primary 
position,  M-N  represents 
a  straight  line  in  the 
third  quadrant.    3/ and  A^  represent  any 
two  points  of  the  line  but  in  no  sense 
limit  the  length  of  the  line.    The  hori- 
zontal projection  of  M  is  m^p  and  the 
vertical  projection  of  M  is    m".     The 
horizontal  projection  of   A^  is   n^j,  and 
the  vertical  projection  of  A^is  n".    The  horizontal  projection  of  the 
line  is  then  rrij-nj^,  and  the  vertical  projection  of  the  line  is  m"-n". 

The  projection  of  a  straight  line,  then,  may  be  found  by  finding 
the  projections  of  two  of  its  points  and  drawing  a  straight  line 
through  these  two  projections.  This  will  be  true  whatever  the 
quadrant  occupied  by  the  line. 


Fig.  6 


12 


DESCRIPTIVE  GEOMETRY 


It  will  be  noticed  that  when  the  horizontal  and  vertical  pro- 
jections of  a  straight  line  are  given,  the  line  is  in  general  definitely 
determined,  for  the  horizontal  and  vertical  projecting  planes  deter- 
mined by  the  projections  of  the  line  intersect  in  the  only  line  which 
can  have  its  horizontal  and  vertical  projections  in  these  lines. 

It  will  be  also  noticed  that  the  two  projections  of  a  line  deter- 
mine the  position  of  the  line  with  reference  to  H  and  V.  For 
example,  in  Fig.  6  the  vertical  projection  of  the  line  shows  that 
the  line  slopes  downward  to  the  right,  the  horizontal  projection 
shows  that  the  line  approaches  Fas  it  extends  toward  the  right, 


Fig.  7 


Fig.  8 


and  the  two  projections  together  show  that  the  line  slopes  down- 
ward toward  V  to  the  right. 

24.  Representation  of  the  Straight  Line  upon  H  and  V  in  their 
Position  of  Coincidence.  In  Fig.  7  the  projections  'mj-rtj^  and 
m"-n"  are  those  previously  found  in  Fig.  6,  and  represent  a 
straight  line  in  the  third  quadrant. 

If  the  plane  V  remains  stationary  and  the  plane  //  be  revolved 
as  previously  directed,  the  points  m"  and  n"  will  remain  stationary, 
while  the  points  ?/?,,  and  w,,  will  revolve  and  fall  at  the  points  m, 
and  n,  respectively.  The  line  m-n,,  then,  will  represent  the 
horizontal  projection  of  the  line  M-N  after  H  has  been  revolved, 
and  the  result  may  be  expressed  as  shown  in  Fig.  8. 


THE  POINT,  LINE,  AND  PLANE 


13 


It  will  be  easily  seen  that  we  shall  obtain  the  same  result  if  we 
allow  the  horizontal  plane  of  projection  to  remain  stationary  and 
revolve  Fas  directed  in  Section  14. 

It  will  be  observed  that  the  two  projections  of  the  line  in  Fig.  8 
bear  the  same  relation  to  G-L  as  they  did  in  Fig.  6,  and  therefore 
reveal  just  as  much  in  regard  to  the  position  of  the  line  itself. 

25.  Projections  of  Straight  Lines  occupying  Various  Positions  with 
Reference  to  H  and  F.  In  Fig.  9,  M-N  represents  a  straight  line 
in  the  third  quadrant  perpendicular  to  H.  The  horizontal  projec- 
tion is  a  point  m,,  n,.  The  vertical  projection  is  a  straight  line 
m^-n'  perpendicular  to  G-L  and  parallel  to  M-N.    That  portion 


m^ 


n' 


\ h 


d^p 


Fig.  10 


of  the  vertical  projection  included  between  w!  and  'n!  is  equal  to 
the  distance  between  M  and  N. 

In  the  same  figure  0-F  represents  a  straight  line  in  the  third 
quadrant  perpendicular  to  V.  The  vertical  projection  is  a  point 
o\  p'.  The  horizontal  projection  is  a  straight  line  o-p^  perpen- 
dicular to  G-L  and  parallel  to  0-P.  That  portion  of  the  horizontal 
projection  included  between  o,  and  p,  is  equal  to  the  distance 
between  0  and  P. 

Fig.  10  shows  how  the  projections  of  Fig.  9  will  appear  after 
H  and.F  take  their  position  of  coincidence. 

In  Fig.  11,  Jf-xY  represents  a  straight  line  in  the  third  quadrant, 
parallel  to  //  but  oblique  to  V.  The  vertical  projection  rnl-n'  is 
parallel  to  G-L  and  at  the  same  distance  below  G-L  that  M-N  is 


14 


DESCEIPTIVE   GEOMETRY 


below  H.  The  horizontal  projection  m-Uj  is  parallel  to  M-N^  and 
makes  the  same  angle  with  G-L  that  M-N  makes  with  V.  That 
portion  of  the  horizontal  projection  nicluded  between  m,  and  n^  is 
equal  to  the  distance  between  31  and  N. 

In  the  same  figure  0-P  represents  a  straight  line  in  the  third 
quadrant,  parallel  to  Fbut  oblique  to  H.  The  horizontal  projection 
0,-pj  is  parallel  to  G—L  and  at  the  same  distance  back  of  G-L 
that  0-P  is  back  of  V.    The  vertical  projection  o^-p'  is  parallel 


Fig.  11 


Fig.  12 


Fig.  13 


Fig.  14 


to  0-P  and  makes  the  same  angle  with  G-L  that  0-P  makes  with 
H.  That  portion  of  the  vertical  projection  included  between  o'  and 
p^  is  equal  to  the  distance  between  0  and  P. 

Fig.  12  shows  how  the  projections  of  Fig.  11  will  appear  after 
^  and  F  take  their  position. of  coincidence. 

In  Fig.  13,  7lf-iV  represents  a  straight  line  in  the  second  quad- 
rant and  oblique  to  both  7/  and  V, 


THE  P01]ST,  LINE,   AND  PLAXE 


15 


Fig.  14  shows  how  the  projections  of  Fig.  13  will  appear  after 
H  and  V  take  their  position  of  coincidence. 

In  Fig.  15,  3f-N  represents  a  straight  line  located  by  the  point 
M\n  the  third  quadrant  and  by  the  point  N  in  the  first  quadrant. 
The  line  pierces  F  at  a'  and  pierces  if  at  5,.  The  section  M-h^  of 
the  line  is  in  the  third  quadrant,  the  section  h-a'  is  in  the  second 
quadrant,  and  the  section  a'-N  is  in  the  first  quadrant. 


Fig.  17 


Fig.  18 


Fig.  16  shows  how  the  projections  of  Fig.  15  will  appear  after 
H  and  V  take  their  position  of  coincidence. 

In  Fig.  17,  Jf-A^  represents  a  straight  line  located  by  the  point 
M  in  the  fourth  quadrant  and  by  the  point  A^  in  the  second  quad- 
rant. The  line  pierces  H  at  5,,  runs  through  the  first  quadrant, 
and  pierces  F  at  a^ 

Fig.  18  shows  how  the  projections  of  Fig.  IT  will  appear  after 
H  and  V  take  their  position  of  coincidence. 


16 


DESCEIPTIVE  GEOMETRY 


In  Fig.  19,  Jf-iV  represents  a  straight  line  in  a  profile  plane  and 
located  by  a  point  M  in  the  first  quadrant  and  by  a  point  JY  in  the 
third  quadrant. 

Fig.  20  shows  how  the  projections  of  Fig.  19  will  appear  after 
IT  and  V  take  their  position  of  coincidence. 

26.  Observations.  When  a  straight  line  is  parallel  to  a  plane  of 
projection,  the  projection  upon  this  plane  of  any  definite  portion 
of  the  line  will  be  equal  in  length  to  the  portion  in  question. 


mr 


n, 


Fig.  19 


Fig.  20 


If  a  straight  line  is  situated  in  a  plane  of  projection,  the  line 
itself  is  its  own  projection  upon  this  plane. 

If  a  straight  line  is  situated  in  H^  its  horizontal  projection  is  the 
line  itself  and  its  vertical  projection  is  in  G-L. 

If  a  straight  line  is  situated  in  F,  its  vertical  projection  is  the 
line  itself  and  its  horizontal  projection  is  in  G-L. 

If  a  straight  line  is  parallel  to  H  and  oblique  to  F,  its  vertical 
projection  will  be  parallel  to  G-L  and  its  horizontal  projection 
will  make  the  same  angle  with  G-L  that  the  line  itself  makes 
with  V. 

If  a  straight  line  is  parallel  to  Fand  oblique  to  ^,  its  horizontal 
projection  will  be  parallel  to  G-L  and  its  vertical  projection 
will  make  the  same  angle  with  G-L  that  the  line  itself  makes 
with  H. 

If  a  straight  line  is  in  a  profile  plane,  its  projecting  planes 
coincide  and  its  projections  are  perpendicular  to  G-L. 


THE  POINT,   LINE,   AND  PLANE 


17 


Fig.  21 


27.  To  assume  a  Straight  Line  at  Random.  To  assume  a  straight 
line  at  random,  we  may,  within  certain  limitations,  draw  its  two 
projections  at  random. 

28.  To  assume  a  Point  upon  a  Straight  Line.  In  Fig.  21  let 
m-Uf  and  m'—n'  represent  the  horizontal  and  vertical  projections 
of  a  straight  line  in  the  third  quadrant. 
Assume  one  of  the  projections  of  the  point, 
say  the  horizontal,  o,,  anywhere  on  the 
line  m-Tif.  Through  o^  draw  a  straight 
line  perpendicular  to  G-L  and  note  its 
intersection  o'  with  m'-7i'.  The  point  o' 
is  the  vertical  projection  of  the  required 
point. 

If  a  point  is  situated  on  a  line,  its  hori- 
zontal projection  must  be  on  the  horizontal  projection  of  the  line, 
and  its  vertical  projection  must  be  on  the  vertical  projection  of 
the  line. 

29.  To  assume  Two  Straight  Lines  which  intersect.  See  Fig.  22. 
Draw  at  random  the  two  projections,  m,-w,  and  m'-n',  of  one  of 
the  lines.  Assume  upon  this  line  any  point  as  P.  Through  p, 
and  p'  respectively  draw  at  random  the  horizontal  and  vertical 

projections  of  the  other  line. 

30.  The  Projections  of  Parallel 
Straight  Lines.  The  projecting  planes 
of  parallel  straight  lines  are  necessarily 
parallel  whatever  the  plane  of  projec- 
tion ;  therefore  the  intersections  of  these 
planes  with  the  plane  of  projection,  or 
in  other  words  the  projections  of  the 
lines,  must  be  parallel. 

31.  The  Projections  of  Straight  Lines 
which  are  Perpendicular  to  Each  Other.  The  projections  of  two 
straight  lines  which  are  perpendicular  to  each  other  are  perpen- 
dicular to  each  other  only  when  one  or  both  of  the  lines  are 
parallel  to  the  plane  of  projection,  since  it  is  only  under  these  con- 
ditions that  the  projecting  planes  of  the  lines  are  perpendicular 
to  each  other. 


Fig.  22 


18 


DESCRIPTIVE  CxEOMETRY 


32.  Problem  8.  Draw  the  two  projections  of  a  straight  line  situated 
in  the  second  quadrant  parallel  to  H,  oblique  to  F,  and  2  units 
from  H. 

33.  Problem  9.  Draw  the  two  projections  of  a  straight  line  situated 
in  the  third  quadrant  perpendicular  to  H  and  3  units  from  V. 

34.  Problem  10.  Given  a  point  Jf,  6  units  hack  of  V  and  4  units 
below  //,  also  a  point  iV,  4  units  in  front  of  V  and  4  units  above  H ; 
required  to  draw  the  projections  of  the  straight  line  located  by  the 
points  M  and  N. 

35.  Problem  11.  Draw  the  two  projections  of  a  straight  line  situated 
in  the  fourth  quadrant  parallel  to  H  and  V  and  equidistant  from  H 
and  V, 

36.  Problem  12.  Draw  the  two  projections  of  a  straight  line  situated 
in  the  fourth  quadrant  perpendicular  to  V  and  4  units  from  H. 


Fig.  23 


Fig.  24 


37.  Representation  of  the  Plane.  The  line  in  which  a  plane  inter- 
sects H  is  called  the  horizontal  trace  of  the  plane.  The  line  in 
which  a  plane  intersects  V  is  called  the  vertical  trace  of  the 
plane. 

If  either  trace  of  a  plane  intersects  G-L^  the  other  trace  must 
intersect  G-L  at  the  same  point,  otherwise  G-L  would  intersect 
the  plane  in  two  points. 

If  either  trace  of  a  plane  is  parallel  to  G-L,  the  other  trace 
must  also  be  parallel  to  G-L,  otherwise  G-L  would  intersect  the 
plane  while  it  was  parallel  to  a  straight  line  in  the  plane. 

As  a  plane  is  completely  determined  either  by  two  intersecting 
straight  lines  or  by  two  parallel  straight  lines,  it  is  evident  that  a 
plane  is  definitely  located  bv  its  traces. 


THE  POINT,  LINE,   AND  PLANE 


19 


In  Fig.  23,  S-8^  represents  the  horizontal  trace,  and  S-s'  repre- 
sents the  yertical  trace  of  a  plane  S  which  is  oblique  to  both  H 
and  V.  The  plane  is  supposed  to  extend  without  limit  in  all 
directions,  and  therefore  passes  through  all  the  quadrants,  although 
in  the  diagram  only  that  portion  of  the  plane  which  lies  in  the 
third  quadrant  is  represented. 

Fig.  24  shows  how  the  traces  of  the  plane  in  Fig.  23  will 
appear  after  H  and  V  take  their  position  of  coincidence. 

It  will  be  observed  that  the  traces  S-s^  and  S-s'  make  the  same 
angle  with  G-L  after  revolution  as  before,  and  therefore  reveal 
just  as  much  in  regard  to  the  location  of  the  plane. 

38.  Representation  of  the  Plane  by  Two  Intersecting  Straight 
Lines  in  Space.    Since  any  two  intersecting  straight  lines  or  any 


Fig.  25 


Fig.  26 


two  parallel  straight  lines  may  be  used  to  determine  a  plane,  we 
may  represent  a  plane  by  giving  the  projections  of  such  lines. 

By  producing  these  lines  until  they  intersect  H  and  V  we  shall 
find  points  in  the  horizontal  and  vertical  traces  of  the  plane,  if 
such  traces  are  needed. 

39.  Representation  of  Planes  occupying  Various  Positions  with 
Reference  to  H  and  F.  Fig.  25  represents  a  plane  in  the  third 
quadrant,  perpendicular  to  R  but  oblique  to  V.  Under  these  con- 
ditions the  vertical  trace  must  be  perpendicular  to  G-L  and  the 
horizontal  trace  must  make  the  same  angle  with  G-L  that  the 
plane  makes  with   V. 

Fig.  26  shows  how  the  traces  of  the  plane  in  Fig.  25  will  appear 
after  H  and  V  take  their  position  of  coincidence,  no  change  taking 
place  in  the  relation  of  the  traces  to  G-L. 


20 


DESCRIPTIVE  GEOMETKY 


Fig.  27  represents  a  plane  in  the  third  quadrant,  perpendicular 
to  V  but  oblique  to  H.  Under  these  conditions  the  horizontal  trace 
must  be  perpendicular  to  G-L  and  the  vertical  trace  must  make 
the  same  angle  with  G-L  that  the  plane  makes  with  H. 

Fig.  28  shows  how  the  traces  of  the  plane  in  Fig.  27  will  appear 
after  H  and  V  take  their  position  of  coincidence. 


Fig.  27 


Fig.  28 


5- 


8- 


FiG.  29 


Fig.  30 


Fig.  29  represents  a  plane  in  the  third  quadrant  and  parallel  to 
G-L.    Its  traces  then  must  both  be  parallel  to  G-L. 

Fig.  30  shows  how  the  traces  of  the  plane  in  Fig.  29  will  appear 
after  H  and  V  take  their  position  of  coincidence. 

In  all  the  foregoing  cases,  in  order  that  the  diagrams  may  be 
more  easily  understood,  only  limited  portions  of  the  planes  have 
been  taken  into  consideration ;    but  planes  must  be  regarded  as 


THE  POINT,  LINE,   AND  PLANE 


21 


infinite  in  extent  and  in  no  way  limited  by  their  traces  or  by  any 
lines  which  locate  them. 

When  a  plane  is  oblique  to  G-L^  as  the  plane  S  in  Figs.  31  and  32, 
it  passes  through  all  the  quadrants.  Its  horizontal  trace  extends  both 
in  front  and  back  of  G-L  and  its  vertical  trace  extends  both  above 
and  below  G-L.  The  traces  S-s^^  and  S-s"  limit  that  portion  of  the 
plane  in  the  first  quadrant,  the  traces  S-s^  and  S-s^'  limit  that  por- 
tion of  the  plane  in  the  second  quadrant,  the  traces  S-s^  and  S-s' 
limit  that  portion  of  the  plane  in  the  third  quadrant,  and  the  traces 
S-s  J  J  and  S-s'  limit  that  portion  of  the  plane  in  the  fourth  quadrant. 


Fig.  31 


Fig.  32 


40.  To  assume  a  Plane  at  Random.  Assume  the  two  traces  at 
random  so  long  as  they  either  intersect  G-L  at  the  same  point  or 
are  both  parallel  to  G-L. 

41.  To  assume  a  Series  of  Parallel  Planes.  Since  the  intersec- 
tions of  a  series  of  parallel  planes  by  an  oblique  plane  is  a  series 
of  parallel  straight  lines,  it  follows  that  the  horizontal  traces  of 
parallel  planes  must  be  parallel;  also  that  their  vertical  traces 
must  be  parallel. 

42.  To  assume  a  Straight  Line  in  a  Plane. 

Principle.  If  a  straight  line  in  a  plane  pierces  H  it  must  pierce 
it  in  the  horizontal  trace  of  the  plane,  and  if  it  pierces  V  it  must 
pierce  it  in  the  vertical  trace  of  the  plane ;  for  the  horizontal  trace 
of  the  plane  is  the  only  line  in  common  between  the  plane  and  H^ 
and  the  vertical  trace  of  the  plane  is  the  only  line  in  common 
between  the  plane  and  V, 


22  DESCRIPTIVE   GEOMETRY 

Construction,    In  Fig.  33  let  S-s^  represent  the  horizontal  trace 

and  let  S-s'  represent  the  vertical  trace  of  a  plane  oblique  to  G-L. 

Assume  any  point,  as  Ji,  on  the  horizontal  trace.    The  point  m^ 

is  the  horizontal  projection  of  this  point,  and  the  vertical  projection 

must  be  in  G—L  at  m'. 

Assume  another  point,  as  iV",  anywhere  in  the  vertical  trace.  The 
point  n'  is  the  vertical  projection  of  this  point,  and  the  horizontal 
projection  must  be  in  G-L  at  n,. 

Since  M  is  a  point  in  the  horizontal  trace  of  the  plane  S^  it  must 
be  a  point  in  the  plane  ;  and  since  iV^is  a  point  in  the  vertical  trace  of 

^         the  plane  S^  it  must  be  a  point 
^^*^^^         in  the  plane.    The  line  3I-N  is 
^^^.^y^^   :  therefore  a  line  of  the  plane. 

^^^-^^^'^^^/^  \   ,  Prhieiple.    If    through    any 

^         "^^^'^i    — ^^^^^  '      ^    point  in  a  plane  a  straight  line 

J>>^::^^^  is   drawn    parallel    to    another 

^^^^^^  straight    line    already    in    the 

^""^^"^^  plane,  the  line  so  drawn  is  also 

^^^-  ^^  in  the  plane. 

Principle.    If  a  straight  line  is  in  a  plane  and  is  parallel  to  ^, 

it  will  be  parallel  to  the  horizontal  trace  of  the  plane,  and  its 

horizontal  projection  will   be  parallel  to  the  horizontal  trace  of 

the  plane. 

Principle.  If  a  straight  line  is  in  a  plane  and  is  parallel  to  F,  it 
will  be  parallel  to  the  vertical  trace  of  the  plane,  and  its  vertical 
projection  will  be  parallel  to  the  vertical  trace  of  the  plane. 

43.  To  assume  a  Point  in  a  Plane.  First,  by  Section  42,  assume 
a  straight  line  in  the  plane ;  then  by  Section  28  assume  a  point 
upon  this  line. 

44.  Proposition  1.  If  a  straight  line  is  perpendicular  to  a  plane, 
its  horizontal  projection  will  be  perpendicular  to  the  horizontal  trace 
of  the  plane,  and  its  vertical  projection  will  he  perpe7idicular  to  the 
vertical  trace  of  the  plane. 

Proof.  Since  the  line  is  perpendicular  to  the  given  plane,  the 
horizontal  projecting  plane  of  the  line  must  be  perpendicular  to 
the  given  plane.  The  horizontal  projecting  plane  of  the  line  is 
also  perpendicular  to  H,  and  therefore  since  it  is  perpendicular 


THE  POINT,  LINE,  AND  PLANE  23 

both  to  the  given  plane  and  to  H^  it  must  be  perpendicular  to 
their  intersection,  which  is  the  horizontal  trace  of  the  given  plane. 
Since  the  horizontal  trace  of  the  given  plane  is  perpendicular  to 
the  horizontal  projecting  plane  of  the  given  line,  it  must  be  per- 
pendicular to  any  line  of  the  horizontal  projecting  plane  passing 
through  its  foot,  as  the  horizontal  projection  of  the  given  line. 

By  a  similar  course  of  reasoning  in  connection  with  the  ver- 
tical projecting  plane  of  the  given  line  we  may  prove  that  the 
vertical  projection  of  the  given  line  must  be  perpendicular  to 
the  vertical  trace  of  the  given  plane. 

45.  Proposition  2.  If  the  two  projections  of  a  straight  line  are  per- 
pendicular respectively  to  the  two  traces  of  a  given  plane,  the  line 
itself  will,  in  general,  be  perpendicular  to  the  plane. 

Proof.  The  horizontal  projecting  plane  of  the  line  is  perpen- 
dicular to  the  horizontal  trace  of  the  given  plane  and  therefore 
perpendicular  to  the  plane.  The  vertical  projecting  plane  of  the 
line  is  perpendicular  to  the  vertical  trace  of  the  given  plane  and 
therefore  perpendicular  to  the  plane.  Since  the  horizontal  and 
vertical  projecting  planes  of  the  line  are  both  perpendicular  to  the 
given  plane,  their  intersection,  which  is  the  line  itself,  must  be 
perpendicular  to  the  plane. 

46.  Problem  13.  Draw  the  traces  of  a  plane  which  is  perpendicular 
to  H  and  makes  an  angle  of  45  degrees  with  V. 

47.  Problem  14.  Draw  the  traces  of  a  plane  which  is  perpendicular 
to  V  and  makes  an  angle  of  60  degrees  with  IT. 

48.  Problem  15.  Given  a  plane  which  makes  an  angle  o/*30  degrees 
with  H,  and  whose  horizontal  trace  is  parallel  to  G—L  and  4  units 
hack  of  V ;  required  to  draw  the  vertical  trace. 

49.  Problem  16.  Bratv  the  traces  of  a  plane  parallel  to  the  plane 
determined  in  Problem  15. 

50.  Problem  17.  Given  a  plane  which  is  oblique  to  G-L,  whose 
horizontal  trace  is  back  of  F,  and  whose  vertical  trace  is  above  H ; 
required  to  draiv  the  projections  of  a  straight  line  in  this  plane. 

51.  Problem  18.  Given  a  plane  which  is  parallel  to  G-L  ;  required 
to  draw  the  projections  of  a  point  in  this  plane. 


CHAPTER  III 


SUPPLEMENTARY  PLANES   OF  PROJECTION 

52.  Supplementary  Planes  of  Projection.  While  H  and  Fare  the 
principal  planes  of  projection,  projection  may  be  made  upon  any 
plane  which  we  may  choose  for  that  purpose.  Any  plane  other 
than  H  OT  V  which  is  used  as  a  plane  of  projection  is  spoken  of  as 
a  supplementary  plane  of  projection. 

The  most  common  supplementary  plane  of  projection  is  a  profile 
plane  of  projection,  —  a  plane  perpendicular  to  both  K  and  V  and 
consequently  perpendicular  to  G-L. 

53.  Representation  of  the  Point  upon  a  Profile  Plane  of  Projec- 
tion.   In  Fig.   34,  which  is  a  pictorial  drawing,   let  the  points 

ili,  N,  0,  and  P  have 
the  same  location  with 
reference  to  IT  and  Fas 
in  Fig.  1,  and  let  the 
plane  P  represent  a 
profile  plane  of  pro- 
jection and  situated  to 
the  right  of  the  given 
points. 

According  to  Sec- 
tion 5,  the  profile  pro- 
jection of  M,  or  the 
projection  of  M  upon 
P,  is  MF;  the  profile 

projection  of  J^  is  NF;  the  profile  projection  of  0  is  OF;  and  the 

profile  projection  of  P  is  FF. 

Here  it  will  be  observed  that  the  distance  of  the  point  itself 

from  V  is  equal  to  the  distance  of  its  profile  projection  from  F", 

and  that  the  distance  of  the  point  itself  from  IT  is  equal  to  the 

distance  of  its  profile  projection  from  ZT. 

24 


Fig.  34 


SUPPLEMENTARY  PLANES  OF  PROJECTION 


25 


771: 


W- 


It  will  also  be  observed  that  when  the  projections  of  a  point 
upon  IT,  F,  and  F  are  given,  the  position  of  the  point  with  reference 
to  these  three  planes  of  projection  is  definitely  determined. 

Revolve  the  plane  F  about  its  vertical  trace  as  an  axis  until 
that  portion  of  the  plane  F  in  front  of  V  falls  on  V  to  the  left  of 
the  axis,  and  that  portion  of  F  back  of  V  falls  on  V  to  the  right 
of  the  axis. 

Each  profile  projection  will  move  in  a  plane  parallel  to  IT,  and 
in  the  arc  of  a  circle  with  center  in  the  axis.  MF  will  fall  upon  V 
at  a  point  as  far  to  the  left  IR 

of  the  axis  as  31  was  origi- 
nally in  front  of  F,  and  as  far 

above  G-L  as  M  was  origi-  ;  I  jipv 

nally  above  If. 

NF  will  fall  upon  V  at  Si    , 
point  as  far  to  the  right  of 

the  axis  as  N  was  originally  ]  ' "        b]^ 

behind  F,  and  as  far  above  P^- 

G-L   as   N  was    originally 

above  H,    OF  will  fall  upon 

r  at  a  point  as  far  to  the 

right  of  the  axis  as  0  was 

originally  behind  F,  and  as  far  below   G-L  as   0  was  originally 

below  H.    FF  will  fall  upon  F  at  a  point  as  far  to  the  leit  of  the 

axis  as  the  point  F  was  originally  in  front  of  F,  and  as  far  below 

G-L  as  the  point  F  was  originally  below  H. 

After  the  profile  plane  has  been  thus  revolved,  the  profile  pro- 
jections of  il/,  lY,  0,  and  F  will  appear  as  shown  in  Fig.  35. 

This  is  the  picture  of  the  four  points  which  the  observer 
would  have  if  he  stood  at  an  infinite  distance  to  the  right  and 
looked  along  visual  lines  parallel  to  G-L.  H  would  appear  as 
a  horizontal  line  corresponding  with  G-L  in  Fig.  35,  and  the 
vertical  plane  would  appear  as  a  vertical  line  corresponding  with 


P' 


Fig.  35 


P, 


-P 


Here  points  in  the  first  quadrant  have  their  profile  projections 
to  the  left  of  the  vertical  line  and  above  the  horizontal  line;  points 
in  the  second  quadrant  have  their  profile  projections  to  the  right 


26 


DESCRIPTIVE  GEOMETRY 


of  tlie  vertical  line  and  above  the  horizontal  line;  points  in  the 
third  quadrant  have  their  profile  projections  to  the  right  of  the 
vertical  line  and  below  the  horizontal  line;  points  in  the  fourth 
quadrant  have  their  profile  projections  to  the  left  of  the  verti- 
cal line  and  below  the  horizontal  line. 

Distances  of  points  in  front  or  back  of  V  are  indicated  by  the 
distances  of  their  profile  projections  to  the  left  or  right  of  the 
vertical  line.  Distances  of  points  above  or  below  H  are  indicated 
by  the  distances  of  their  profile  projections  above  or  below  the 
horizontal  line. 

54.  Representation  of  the  Straight  Line  upon  a  Profile  Plane  of 
Projection.  The  profile  projection  of  a  straight  line  is  determined 
by  the  profile  projections  of  two  points  of  the  line. 


Fig.  36 


In  Fig.  36,  which  is  a  pictorial  drawing,  let  M-N  represent  the 
given  line  and  let  the  plane  P  represent  the  profile  plane  of 
projection. 

The  profile  projection  of  M  is  MP^  and  the  profile  projection  of 
N  is  NP,    The  line  MP-NP  is  the  profile  projection  of  3I-N. 

Now  if  we  revolve  the  profile  phane  of  projection  about  its  verti- 
cal trace  P-p'  until  it  coincides  with  F,  MP  will  fall  at  mp^^  and 
NP  will  fall  at  np^\  The  line  mp^-np^'  is  the  profile  projection  of 
M-N  after  the  revolution. 


SUPPLEMENTARY  PLANES  OF  PROJECTION 


27 


Fig.  37  shows  how  the  projections  of  Fig.  36  will  appear  after 
the  planes  H^  F,  and  P  have  taken  their  position  of  coincidence. 


Fig.  38 


Fig.  38  shows  the  projections  upon  H,  V,  and  P  of  a  straight 
line  ilif-iV  running  from  a  point  in  the  fourth  quadrant  to  a  point  in 
the  second  quadrant. 

Fig.  39  shows  the  projections  upon  H,  F,  and  P  of  a  straight 
line  M-H  in  a  profile  plane. 


28 


DESCKIPTIVE  GEOMETRY 


As  profile  planes  are  parallel  the  distance  mp^-np^  in  Fig.  39 
must  be  equal  to  the  distance  M-N. 

55.  Other  Supplementary  Planes  of   Projection.    Supplementary 
planes  of  projection  may  be  assumed  in  any  position  whatever. 

Fig.  40,  which  is  a  picto- 
rial drawing,  represents  a 
supplementary  plane  of 
projection  U,  assumed  par- 
allel to  the  horizontal  pro- 
jecting plane  of  the  line 
M-N. 

According  to  Section  5 
the  supplementary  projec- 
tion of  M  in  this  case  is 
MU  and  the  supplemen- 
tary projection  of  N  is 
NU,  Therefore  the  sup- 
plementary projection  of  M-N  is  MU-NU. 

It  will  be  observed  that  because  the  supplementary  plane  of 
projection  is  perpendicular  to  H^  the  supplementary  projection  of 
any  point,  as  Jf,  will  fall  at 
the  same  distance  from  the 
horizontal  trace  U-u,  of 
the  supplementary  plane 
as  the  point  M  is  from  H^ 
or,  what  is  the  same  thing, 
as  far  as  the  vertical  pro- 
jection m'  is  from  G-L. 

It  will  also  be  observed 
that  the  lines  MU-mu^ 
and  rrij-mUj  are  both  per- 
pendicular to  U-u,  at  the 
same  point. 

Now  if  the  supplemen- 
tary plane  U  be  revolved 

about  its  horizontal  trace  as  an  axis  until  it  coincides  with  //, 
MU  will  fall    at    7nu, 


upon   a   straight   line   drawn   through 


m, 


SUPPLEMENTARY  PLANES   OF  PROJECTION 


29 


Fig.  41 


perpendicular  to  U-u,  and  at  a  distance  from  U-u,  equal  to  the 

distance  of  M  from  H.    For  the  same 

reason  NU  will  fall  at  nu^j^  and  the  line 

mujj-nujj  will   represent   in    revolved 

position  the  supplementary  projection 

of  the  line  M-N. 

Fig.  41  shows  how  the  projections 

of  Fig.  40  will  appear  after  the  planes 

H^  V,  and  U  have  taken  their  position 

of  coincidence. 

Fig.  42  represents  a  supplementary 

plane  of  projection  assumed  parallel  to 

the  vertical  projecting  plane  of  a  line 

M-JSF.    This  plane  is  perpendicular  to 

V  and  its  vertical  trace  is  U-u',    The 

supplementary  projection  of  M  is  MUy 

the  supplementary  projection  of  N  is 

NU,  and  the  supplementary  projection 

of  the  line  is  MU-NU. 

It  will  be  observed  that  the  supplementary  projection  of  any 

point,  as  Jf,  upon  this 
plane  is  at  the  same 
distance  from  V  as  the 
point  itself. 

It  will  also  be  ob- 
served that  the  lines 
MTJ-mvJ  and  unJ-mv!  are 
both  perpendicular  to 
TJ-u^  at  the  same  point. 
Now  if  the  supple- 
mentary plane  TJ  be  re- 
volved about  its  vertical 
trace  as  an  axis  until  it 
coincides  with  F,  MU 
will  fall  at  mv)^  upon 
a    straight    line    drawn 

through  7?z'  perpendicular  to  XJ-n!  and  at  a  distance  from  JJ-v! 


Fig-  42 


30 


DESCRIPTIVE  GEOMETRY 


equal  to  the  distance  of  M  from  V.  For  the  same  reason  NU 
will  fall  at  nu^,  and  the  line  mu^^-nic^  will  represent  in  revolved 
position  the  supplementary  projection  of  the  line  M-N. 

Fig.  43  shows  how  the  projections  of  Fig.  42  will  appear  after 
the  planes  //,  F,  and  U  have  taken  their  position  of  coincidence. 

56.  Revolution  of  the  Supplementary  Planes.  After  projection 
has  been  made  upon  a  supplementary  plane  of  projection  the  plane 
is  revolved  about  either  its  horizontal  trace  or  its  vertical  trace 

until  coincident  with  the  correspond- 
ing plane  of  projection. 

The  revolution  is  always  made  in 
such  a  way  that  that  portion  of  the 
supplementary  plane  above  R  or  in 
front  of  V,  according  as  the  axis  is 
in  H  or  in  V,  shall  move  toward  the 
principal  projections  already  made 
upon  //  or  F,  and  that  that  portion 
of  the  supplementary  plane  of  pro- 
jection below  H  or  back  of  V  shall 
move  in  the  opposite  direction  or 
away  from  the  principal  projections 
on  H  or  V. 

Illustration  of  this  practice  may 
be  seen  in  Fig.  36,  where  the  sup- 
plementary projection  of  M-N  upon 
P  back  of  V  is  revolved  to  the  right,  or  away  from  the  projection 
m'-n' ',  also  in  Fig.  38,  where  that  portion  of  the  supplementary 
plane  P  in  front  of  V  is  revolved  toward  the  left  and  that  portion 
back  of  V  is  revolved  toward  the  right,  the  former  toward  and  the 
latter  away  from  the  projection  m'-n' ;  again  in  Fig.  40,  where  that 
portion  of  the  supplementary  plane  U  below  //  is  revolved  toward 
the  right,  or  away  from  the  projection  m-n,. 

In  case  the  supplementary  plane  is  on  the  left  of  the  magni- 
tude, that  portion  of  the  supplementary  plane  above  //  or  in  front 
of  F,  according  as  the  axis  of  revolution  is  in  H  or  in  F,  is  revolved, 
according  to  the  rule,  toward  the  right,  or  in  a  direction  opposite  to 
that  assumed  by  supplementary  planes  on  the  right  of  the  magnitude. 


Fig.  43 


SUPPLEMENTARY  PLANES  OF  PEOJECTION    31 

57.  Problem  19.  Given  a  straight  line  in  the  third  quadrant  and 
oblique  to  H  and  to  V;  required  the  supplementary  projection  of  the 
line  upon  a  profile  plane  of  projection  assumed  on  the  left  of  the  line. 

58.  Problem  20.  Given  a  straight  line  in  the  second  quadrant  and 
oblique  to  H  and  to  V;  required  the  supplementary  projection  of  the 
line  upon  a  plane  parallel  to  the  horizontal  projecting  plane  of  the 
line  and  assumed  on  the  left  of  the  line.  . 

59.  Problem  21.  Given  a  straight  line  in  the  fourth  quadrant  and 
oblique  to  H  and  to  V ;  required  the  supplementary  projection  of  the 
line  upon  a  plane  parallel  to  the  vertical  projecting  plaiie  of  the  line. 

60.  Problem  22.  Given  a  plane  parallel  to  G-L  but  oblique  to  H 
and  to  V;  required  the  angle  which  the  plane  makes  with  H;  also  the 
angle  which  the  plane  makes  with  V.  Solve  by  use  of  a  supplementary 
profile  plane  of  projection. 

61.  Problem  23.  Given  a  plane  parallel  to  G-L  but  oblique  to 
H  and  to  V;  required  the  traces  of  a  number  of  planes  parallel  to 
the  given  plane.  Solve  by  use  of  a  supplementary  profile  plane  of 
projection. 

62.  Problem  24.  Given  a  plane  parallel  to  G-L  and  passing 
through  the  fourth^  first,  and  second  quadrants ;  required  the  traces 
of  a  plane  parallel  to  the  given  plane  and  passing  through  the  fourth, 
third,  and  second  quadrants.  Solve  by  use  of  a  supplementary  profile 
plane  of  projection. 


CHAPTER  IV 

NOTATION 

63.  To  distinguish  between  a  point  in  space  and  its  projections 
on  H  and  F,  the  point  itself  will  be  represented  by  the  capital 
letter  and  its  projection  by  the  small  letter. 

64.  The  horizontal  and  vertical  projections  of  a  point,  as  M^  will 
be  represented  by  m,  and  mJ  respectively,  and  will  be  read  m  sub 
one  and  m  prime  one  respectively. 

65.  If  a  point,  as  M,  is  made  to  occupy  several  positions  in  the 
same  problem,  its  horizontal  and  vertical  projections  will  be  repre- 
sented in  order  by  w,,  m';  m,,^  m"-,  w,,,,  m'^'-,  etc. 

66.  When  two  or  more  points  are  projected  into  the  same  point, 
the  letters  of  all  the  points  will  be  written  upon  this  common 
projection. 

67.  If  a  point,  as  Jf,  is  revolved  into  H^  its  revolved  position 
will  be  represented  by  m^,  and  the  vertical  projection  of  the  point 
in  this  revolved  position  will  be  represented  by  m^. 

68.  If  a  point,  as  Jf,  is  revolved  into  F,  its  revolved  position 
will  be  represented  by  m^,  and  the  horizontal  projection  of  the 
point  in  this  revolved  position  will  be  represented  by  m^. 

69.  The  horizontal  projection  of  a  vertical  projection,  as  m\  will 
be  represented  by  {m!),. 

70.  The  vertical  projection  of  a  horizontal  projection,  as  w,, 
will  be  represented  by  (w^)'. 

71.  If  a  point,  as  Jf,  is  projected  upon  a  plane,  as  T,  other  than 
H  or  F,  this  projection  will  be  represented  by  JfT,  and  the  hori- 
zontal and  vertical  projections  of  this  projection  will  be  represented 
by  mtf  and  mt'  respectively. 

72.  If  a  point,  as  M^  is  projected  upon  a  plane,  as  P,  and  if  this 
projection  is  projected  upon  another  plane,  as  (>,  the  last  projection 
will  be  represented  by  MPQ^  and  its  horizontal  and  vertical  pro- 
jections will  be  represented  by  mpq,  and  mpq^  respectively. 

32 


NOTATION  33 

73.  If  in  the  last  case  MPQ  should  be  revolved  into  //,  its 
revolved  position  would  be  represented  by  mpqjj ;  if  revolved  into 
F,  its  revolved  position  would  be  represented  by  mpq^, 

74.  A  new  ground  line  will  be  represented  by  either  Gj-L^  or 
G'-L\  according  as  it  lies  in  the  original  H  or  V. 

75.  The  projections  of  given  or  required  lines,  if  visible,  will  be 
represented  by  full  lines. 

76.  The  projections  of  given  or  required  lines,  when  invisible, 
and  the  projections  of  all  construction  lines  will  be  represented  by 
broken  lines  consisting  of  dashes  about  \  inch  in  length  and 
separated  by  very  small  spaces,  thus : 


77.  The  traces  of  given  or  required  planes,  if  visible,  will  be 
represented  by  full  lines. 

78.  The  traces  of  given  or  required  planes,  when  invisible,  and 
the  traces  of  all  construction  planes  will  be  represented  by  mixed 
lines,  thus : 


79.  Lines  connecting  the  projections  of  a  point,  also  the  projec- 
tions of  the  path  in  which  a  point  moves,  will  be  represented  by 
dotted  lines,  thus : 


80.  The  planes  H  and  V  and  all  surfaces  and  solids  used  in  the 
construction  of  a  problem  will  be  regarded  as  transparent. 

All  surfaces  and  solids  constituting  given  or  required  portions 
of  a  problem  will  be  regarded  as  opaque. 


CHAPTER  V 

METHOD   OF  LOCATING  GIVEN   PARTS 

81.  Coordinate  Planes  of  Reference.  Conceive  a  profile  plane  of 
projection  F  perpendicular  to  G-L^  passing  through  the  center 
of  the  drawing  space  and  cutting  the  plane  of  the  drawing  in  a 
straight  line  to  be  known  as  the  axis  of  Y. 

All  distances  in  space  will  be  referred  to  the  planes  P,  F,  and 
H^  and  measured  on  straight  lines  perpendicular  to  them. 

Distances  measured  to  the  right  of  P,  in  front  of  V  and  above 
H^  will  be  considered  plus  distances. 

Distances  measured  to  the  left  of  P,  back  of  V  and  below  H^ 
will  be  considered  minus  distances. 

82.  The  Point.  A  point  will  be  located  by  giving  its  distances 
from  P,  F,  and  H.  The  distance  of  a  point  from  P  is  the  same  as 
the  distance  of  its  horizontal  projection  from  the  horizontal  trace 
of  the  plane  P,  or  as  the  distance  of  its  vertical  projection  from 
the  vertical  trace  of  the  plane  P.  The  distance  of  a  point  from  V 
is  the  same  as  the  distance  of  its  horizontal  projection  from  G—L. 
The  distance  of  a  point  from  H  is  the  same  as  the  distance  of  its 
vertical  projection  from  G-L. 

In  giving  the  position  of  a  point  in  space  its  distance  from  P 
will  be  mentioned  first,  its  distance  from  V  second,  and  its  distance 
from  H  last.  31  =  2,  5,  4  means  that  the  point  31  is  2  units  to  the 
right  of  P,  5  units  in  front  of  F,  and  4  units  above  H ;  that  the 
horizontal  projection  of  31  is  2  units  to  the  right  of  Y  and  5  units 
in  front  of  G-L ;  that  the  vertical  projection  of  31  is  2  units  to  the 
right  of  Y  and  4  units  above  G-L.  The  letters  3f,  ^,  0,  P,  Q, 
and  P  will  be  used  in  connection  with  points. 

83.  The  Straight  Line.  A  straight  line  will  be  located  by  giving 
the  position  of  two  of  its  points.  [Jf  =  —  2,  2,  —  5  ;  iV  =  2,  —  4,  6] 
indicates  a  straight  line  passing  through  the  points  3f  and  JSF  but 
not  necessarily  limited  by  them. 

34 


METHOD  OF  LOC^ATINO  GIVEN  PARTS  35 

Straight  lines  will  be  specified  by  the  letters  of  the  points  used 
in  locating  them ;  for  example,  a  straight  line  passing  through  the 
points  M  and  N  will  be  spoken  of  as  the  line  M-N, 

84.  The  Plane.  Planes  will  be  located  by  their  horizontal  and 
vertical  traces,  that  is,  by  their  intersections  with  H  and  V 
respectively. 

The  traces  will  be  located  by  giving  the  position  of  their  vertex, 
that  is,  their  intersection  on  G-L^  and  the  angles  which  they  make 
with  G-L. 

The  angles  which  the  horizontal  and  vertical  traces  make  with 
G-L  will  always  be  measured,  the  former  clockwise,  the  latter 
contra-clockwise,  starting  from  G-L  on  the  right  of  the  vertex. 

Of  the  two  angles  made  with  G-L,  that  made  by  the  horizontal 
trace  will  be  mentioned  first. 

T  =  4,  30°,  45°  means  that  the  vertex  is  4  units  to  the  right  of  Y\ 
that  the  horizontal  trace  runs  forward  toward  the  right,  making 
an  angle  of  30  degrees  with  G-L ;  and  that  the  vertical  trace 
runs  upward  toward  the  right,  making  an  angle  of  45  degrees 
with   G-L. 

S  =  0°,  4,  3  indicates  that  the  traces  are  parallel  to  G-L ;  that 
the  horizontal  trace  is  4  units  in  front  of  G-L ;  and  that  the  verti- 
cal trace  is  3  units  above  G-L. 

Planes  will  be  specified  by  the  letters  aS^,  T,  ZJ,  and  W.  The 
horizontal  traces  will  be  represented  hj  S-s^,  T-t^,  d-Uf,  W-w^; 
and  the  vertical  traces  will  be  represented  by  S-s\  T-t\  U-u',  and 
W-w',  the  capital  letter  being  placed  at  the  vertex. 

85.  Scale.  Problems  of  this  book  whose  given  parts  are  located 
with  reference  to  three  coordinate  planes  of  projection  have  been 
designed  to  the  scale  of  one  quarter  inch  to  the  unit,  and  in  the 
solution  of  these  problems  it  will  be  found  convenient  to  make  use 
of  tliis  scale. 


CHAPTER  VI 

PROBLEMS  RELATING  TO  THE  POINT,  LINE,  AND  PLANE 

86.  Introductory  Statements.  As  a  rule  the  solution  of  a  prob- 
lem will  be  divided  into  two  parts :  (1)  the  analysis,  or  general 
theory  of  solution,  in  which  a  clear  and  logical  statement  of  the 
method  of  procedure  without  reference  to  any  diagram  will  be 
made;  and  (2)  the  construction,  or  actual  graphic  work  necessary 
in  the  solution,  in  which  the  suggestions  of  the  analysis  will  be 
followed  in  order. 

The  graphic  construction  of  every  problem  in  descriptive 
geometry  should  be  checked. 

A  check  in  drafting  is  an  application  of  some  graphic  process 
by  which  the  accuracy  of  construction  may  be  tested.  No  one  can 
be  absolutely  sure  of  the  accuracy  of  his  results  until  they  have 
been  carefully  tested. 

The  nature  of  the  check  for  any  particular  problem  may  be 
determined  by  a  study  of  the  various  conditions  which  must  be 
satisfied  by  the  processes  of  construction. 

A  problem  may  be  said  to  be  checked  when  the  same  result  has 
been  obtained  by  two  distinct  processes  of  solution. 

87.  Problem  25.  Given  a  straight  line  in  H  and  a  point  in  space  ; 
required  to  revolve  the  point  about  the  line  as  an  axis  into  H. 

Principle.  Revolution  in  descriptive  geometry  is  always  made 
about  a  rectilinear  axis  occupying  a  definite  position. 

Principle.  A  point  is  being  revolved  about  an  axis  when  it 
moves  in  a  plane  perpendicular  to  the  axis  and  retains  a  constant 
distance  from  the  axis.  A  point  situated  in  the  axis  will  not  move 
during  revolution. 

Principle.  When  a  plane  is  revolved  about  an  axis  the  relative 
position  of  magnitudes  in  the  plane  is  not  changed. 

Analysis.  In  Fig.  44,  which  is  a  pictorial  drawing,  let  M-N 
represent  the  line  in  H  and  let  0  represent  the  point  in  space. 

36 


RELATING  TO  POINT,  LINE,  AND  PLANE 


37 


Through  o,,  the  horizontal  projection  of  O,  draw  o,-a,  perpen- 
dicular to  rrij-n,.  This  line  must  be  the  horizontal  trace  of  the 
plane  containing  0  and  perpendicular  to  the  axis,  and  therefore 
the  plane  in  which  0  moves  during  revolution. 

After  revolution  the  point  0  must  fall  somewhere  upon  a-o^ 
produced,  and  at  a  distance  from  the  axis  equal  to  its  original 
distance  from  the  axis. 

This  distance  is  0-a,,  which  is  the  hypotenuse  of  the  right- 
angled  triangle  O-o-a^.  The  base  of  this  triangle  is  equal  to  the 
distance  of  the  horizontal  projection  of  the  point  0  from  the  axis, 
and  the  altitude  is  equal  to  the  distance  of  the  point  0  from  H^ 
or,  what  is  the  same  thing,  the  distance  of  the  vertical  projection 
of  the  point  0  from  G-L.    Therefore  lay  off  from  a^  upon  a-o. 

Ok-. 


Fig.  44 


Fig.  45 


produced,  the  distance- a,-o^  equal  to  a^-O,  The  point  Ojj  will  be 
the  revolved  position  of  0. 

Construction,  Fig.  45  shows  how  the  work  is  done  when  the 
planes  H  and  V  occupy  their  position  of  coincidence.  The  line 
0,-a,  is  drawn  through  o^  perpendicular  to  w -w, ,  and  a^-Ojj  is  made 
equal  to  the  hypotenuse  of  a  right-angled  triangle  whose  bas^  is 
equal  to  o^-a^  and  whose  altitude  is  equal  to  o'-h. 

This  hypotenuse  is  calculated  graphically  in  the  triangle  e-d-e, 
where  c-d  equals  o-a^  and  where  e-e  equals  h-o'. 

It  will  be  noticed  in  Fig.  44  that  if  the  point  0  is  so  situated 
that  its  horizontal  projection  falls  on  m-n^,  the  base  of  the  triangle 
will  vanish  and  the  hypotenuse  will  be  equal  to  the  altitude.  In 
such  a  case  then  the  distance  of  the  point  from  the  axis,  or  the 


38  DESCKIPTIVE  GEOMETKY 

radius  of  the  arc  in  which  the  point  moves,  will  be  equal  to  the 
distance  of  the  vertical  projection  of  the  point  from  G-L. 

88.  Problem  26.  G-iven  the  straight  line  [M=  -6,6,0;  N=  6,  -  2, 
0]  and  the  point  (9  =  0,  6,  5  ;  required  to  revolve  0  about  M—N  into  H. 

89.  Problem  27.  Given  the  straight  line  [M=  -  6,  -  6,  0 ;  iV^=  6,  2, 
0]  and  the  point  0  =  0,  —  6, 4  ;  required  to  revolve  0  about  M—N  into  H. 

90.  Problem  28.  Given  the  straight  line  [Jf  =  —  6,  2,  0 ;  jV=  —  6, 
—  4,  0]  and  the  point  0  =  0,  4,  —  6 ;  required  to  revolve  0  about 
M-N  into  H. 

91.  Problem  29.  Given  a  straight  line  m  V  and  a  point  in  space  ; 
required  to  revolve  the  point  about  the  line  as  an  axis  into  V. 

Analysis.  From  what  has  been  said  in  connection  with  Problem 
25,  it  will  be  evident  that  the  vertical  trace  of  the  plane  of  revo- 
lution must  pass  through  the  vertical  projection  of  the  given  point 
and  must  be  perpendicular  to  the  given  line  ;  also  that  the  radius 
of  the  arc  in  which  the  point  moves  will  be  equal  to  the  hypotenuse 
.^^  of  a  right-angled  triangle  whose  base 

I  is  equal  to  the  distance  of  the  verti- 

j,  cal  projection  of  the  point  from  the 

^  j         j  !  ^   line,  and  whose  altitude  is  equal  to  the 

^"*^^^>^^ !  I  distance  of  the  horizontal  projection 

^d        \7^^-^    !  .  of  the  point  from  G-L. 

/     J  /o'        ^^""*\  Construction.    In  Fig.  46  let  M-N 

^  ^       j ^  represent  the  line  in  V  and  let  0  rep- 

resent the  point  in  space.    Through 
'*^'  o'  draw  <? -a'  perpendicular  to  m^-n\ 

This  must  be  the  vertical  trace  of  the  plane  of  revolution.  Make 
a'-o^  equal  to  the  hypotenuse  of  a  right-angled  triangle  c-d-e^  whose 
base  c-d  is  equal  to  o'-a^  and  whose  altitude  c-e  is  equal  to  b-o^. 

It  will  be  noticed,  as  in  Problem  25,  that  if  the  point  0  is  so 
situated  that  its  vertical  projection  falls  on  m'-n\  the  base  of  the 
triangle  will  vanish  and  the  hypotenuse  will  be  equal  to  the  alti- 
tude. In  such  a  case  then  the  radius  of  the  arc  of  revolution 
will  be  equal  to  the  distance  of  the  horizontal  projection  of  the 
point  from  G-L. 

92.  Problem  30.  Given  the  straight  line  [M=  —  6,  0,  —  5 ;  N=  6,  0, 
4]  and  the  point  0  =  0,  4,  6  ;  required  to  revolve  0  about  M-N  into  V. 


RELATING  TO  POINT,  LINE,  AND  PLANE 


39 


93.  Problem  31.  Given  the  straight  line  [ilf=— 6,0,  —  5;  N=Q, 
0,  —  2]  and  the  point  0  =  0,  —  4,  6  ;  required  to  revolve  0  about 
M-N  into  V. 

94.  Problem  32.  Griven  the  straight  line  [M  =  Q,  0,  6;  AT  =6, 
0,  —  2]  and  the  point  0  =  0,  4,  —  6 ;  required  to  revolve  0  about 
M-N  into  V. 

95.  Problem  33.  Given  a  straight  line  in  space  parallel  to  H^  also 
given  a  point  in  space  ;  required  to  revolve  the  point  about  the  line 
as  an  axis  until  the  plane  of  the  point  and  the  line  is  parallel  to  H. 

Analysis  and  Construction.  In  Fig.  47  let  M-N  represent  the 
given  line  and  let  0  represent  the  given  point. 

As  in  Problem  25,  o-a^  represents  the  horizontal  trace  of  the 
plane  containing  the  point  and  perpendicular  to  the  axis,  and 
therefore  the  plane  in  which  0  moves  during  revolution. 


\9n 

^ 

^ 

\  \ 
!    \ 

)0, 

% 

yfn, 

! 

h 

1 

\o" 

d' 

n' 

Fig.  47 


Fig.  48 


After  revolution  the  horizontal  projection  of  the  point  0  must 
fall  somewhere  upon  o-a,  produced,  and  at  a  distance  from  a,  equal 
to  the  hypotenuse  of  a  right-angled  triangle  whose  base  is  equal 
to  o-a,  and  whose  altitude  is  equal  to  o'-d'^  since  the  axis  is  not 
in  H  but  at  a  distance  b-d'  below  H. 

The  point  o,,  represents  the  horizontal  projection  of  0  in  the 
required  revolved  position,  and  o",  at  the  same  distance  below  H 
as  M-N  and  in  a  straight  line  through  o,^  perpendicular  to  G-L^ 
represents  the  vertical  projection  of  0  in  this  position. 

96.  Problem  34.  Given  a  straight  line  in  space  parallel  to  V,  also 
given  a  point  in  space  ;  required  to  revolve  the  point  about  the  line  as  an 
axis  until  the  plane  of  the  poiiit  and  the  line  is  parallel  to  V.    See  Fig.  48- 


40 


DESCEIPTIVE  GEOMETEY 


97.  Problem  35.  G-iven  a  straight  line  in  space  parallel  to  G-L, 
also  given  a  point  in  space;  required  to  revolve  the  point  about 
the  line  as  an  axis  until  the  plane  of  the  point  and  the  line  is 
parallel  to  H^  and  again  until  the  plane  of  the  point  and  the  line  is 
parallel  to  V. 

98.  Problem  36.  Given  a  straight  line  perpendicular  to  H^  also 
given  a  point  in  space;  required  to  revolve  the  point  about  the  line 
as  an  axis. 

Analysis.  In  Fig.  49,  which  is  a  pictorial  drawing,  let  3I-N 
represent  the  line  and  let  0  {o^,  o')  represent  the  point. 

Since  the  point  0  must  move  in  a  plane  perpendicular  to  M-N 
and  therefore  parallel  to  H^  and  since  the  point  0  must  retain  a 


Fig.  49 


r^ 


Ow 


rni.n, 


dfr-.J 


i  jt/y 


I 


rn' 


n' 

Fig.  50 


constant  distance  from  the  axis  M-N,  the  horizontal  projection  o, 
of  the  point  must  move  in  the  arc  of  a  circle  with  center  at  the 
point  where  the  axis  pierces  H;  and  its  vertical  projection  o'  must 
move  in  a  straight  line  o'-o'"  drawn  through  o'  parallel  to  G-L. 

Now  if  the  point  0  occupy  the  several  positions  indicated  in 
the  diagram,  the  projections  will  fall  at  o,,  o';  o„,  o":  <?,,,,  o'";  etc. 

Construction.  Fig.  5  0  shows  how  the  work  is  done  when  the  planes 
^and  V  occupy  their  position  of  coincidence.  The  line  m'-n'  repre- 
sents the  vertical  projection  of  the  given  line,  and  the  point  m,  repre- 
sents the  horizontal  projection  of  the  same  line.  The  points  o,  and  o' 
represent  the  two  projections  of  the  given  point  in  its  first  position. 


RELATING  TO  POINT,   LINE,  AND  PLANE 


41 


"o,« 


With  m,  as  a  center  and  with  m-Oj  as  a  radius  draw  the  circle 
Of—Of-Offf,  This  circle  represents  the  path  of  the  horizontal  projec- 
tion of  the  point  during  revolution.  Through  o'  draw  the  straight 
line  o'-o'"  parallel  to  G-L.  This  line  represents  the  path  of  the 
vertical  projection  of  the  point  during  revolution. 

Now  when  0  occupies  such  a  position  that  its  horizontal  projec- 
tion falls  at  o„,  its  vertical  projection  will  take  the  position  o"  in 
a  straight  line  through  o^^  perpendicular  to  G-L. 

In  the  same  way  the  two  projections  of  0  in  any  position  of  the 
circuit  may  be  found. 

99.  Problem  37.    G-ivefi  the  straight  line  [M  —  0,  —  6,  —  6 ;  iV=  0, 

—  6,-1]  and  the  point  0  =  —  3,  2,  3 ;  required  to  revolve  0  about 
M-N  as  an  axis  through  arcs  o/  30  degrees  and  160  degrees. 

100.  Problem  38.    Given  the  straight  line  [3f  =  0,  —  2,  6 ;  iV=  0, 

—  2,  1]  and  the  point  0  =  3,  4,  —  2 ;  required  to  revolve  the  point  0 
about  M—N  as  an  axis  through  ares  of 
45  degrees  and  135  degrees. 

101.  Problem  39.  Given  a  straight  line 
perpendicular  to  F,  also  given  a  point  in 
space  ;  required  to  revolve  the  point  about 
the  line  as  an  axis. 

Analysis  and  Construction,  In  Fig.  51 
let  M-N  represent  the  given  line,  and 
let  0  (Op  o')  represent  the  given  point. 
Since  the  point  0  must  move  in  a  plane 
perpendicular  to  M-N  and  therefore  par- 
allel to  F,  and  since  the  point  0  must 
retain  a  constant  distance  from  the  axis 
M-N^  the  vertical  projection  of  the  point,  namely  o^  must  move  in 
the  arc  of  a  circle  with  center  at  the  point  where  the  axis  pierces 
F;  and  its  horizontal  projection  o,  must  move  in  a  straight  line  o-o^, 
drawn  through  o,  parallel  to  G-L.  Now  if  the  point  0  occupy  the 
several  positions  indicated  in  the  diagram,  the  projections  will  fall 
at  Op  (?';  o,p  o"\  0,,,,  o'" ;  etc. 

102.  Problem  40.    Given  the  straight  line  [Jf  =  0,  -  1,  -  6  ;  N=  0, 

—  Q^  —  Q-^  and  the  point  (9  =  —  3,  —  4,  2  ;  required  to  revolve  the  point  0 
about  M-N  as  an  axis  through  arcs  of  30  degrees  and  90  degrees. 


Gf- 


m, 


n. 


\y 


'V 


j 
I 


Fig.  51 


42 


DESCEIPTIVE  GEOMETKY 


Fig.  52 


103.  Problem  41.  Given  the  straight  line  [Jf  =  0,  1,  6;  iV^=  0, 
6,  6]  and  the  point  0  =  3,  4,  —  2 ;  required  to  revolve  the  poi7it  0 
about  M-N  as  an  axis  through  arcs  q/*  60  degrees  and  90  degrees, 

104.  Problem  42.  6^^vew  two  inter- 
secting straight  lines,  one  of  which  is 
perpendicular  to  H  and  the  other  oblique 
to  H;  required  to  revolve  the  latter  about 
the  former  as  an  axis. 

Analysis  and  Construction.  In  Fig. 
52  let  M-N  (m—n^,  m'—n')  represent  the 
line  perpendicular  to  H,  and  let  M-0 
(nif-Of,  m'-o')  represent  the  line  oblique 
to  J7. 

Since  the  point  J/,  in  the  line  31-0, 
is  also  in  the  axis  M-N,  it  will  not 
move  during  revolution. 

Any  other  point,  as  0,  in  31-0  will 
move  as  explained  in  Problem  36. 

When  the  line  M-0  occupies  such  a  position  that  its  horizontal 
projection  falls  at  m-o,,,  the  vertical  projection  of  0  will  fall  at  o", 
and  the  vertical  projection  of  the  line 
in  this  position  will  fall  at  m'-o". 

In  the  same  way  the  two  projections 
of  M-0  in  any  position  of  the  circuit 
may  be  found. 

105.  Problem  43.  Given  two  intersect- 
ing straight  lines  one  of  which  is  perpen- 
dicular to  V  and  the  other  oblique  to  V; 
required  to  revolve  the  latter  about  the 
former  as  an  axis. 

Analysis  and  Construction.  In  Fig.  53 
\QtM-N{m-nj,  m'-n^)  represent  the  line 
perpendicular  to  F,  and  let  M-0  {m^-Oj, 
m'-o')  represent  the  line  oblique  to  V. 

For  the  same  reasons  presented  in  Section  104,  J/,  which  is  a 
point  in  the  axis,  will  remain  stationary,  and  0  will  move  as 
indicated  in  the  diagram. 


Fig.  53 


RELATING  TO  POINT,  LINE,  AND  PLANE  43 

106.  Problem  44.  Draw  the  two  projections  of  a  straight  line  pass- 
ing through  the  point  M  =  —  4,  —  4,  2,  running  parallel  to  V,  and 
making  ayi  angle  of  QO  degrees  with  If. 

107.  Problem  45.  Draw  the  two  projections  of  a  straight  line 
passing  through  the  point  Jf  =  — 4,  3,  —1,  running  parallel  to  H^ 
and  makiyig  an  angle  of  45  degrees  with  V. 

108.  Problem  46.  Given  the  straight  line  [M  =  —  4,  5,  1 ;  iV=  4, 
2,  6] ;  required  to  find  the  true  distance  between  M  and  iV,  and  to 
determine  the  angle  which  the  line  makes  with  H. 

Analysis.  Revolve  the  line  about  an  axis  through  M  perpendicu- 
lar to  H  until  the  line  is  parallel  to  V.  The  vertical  projection  of 
any  portion  of  the  line  in  this  new  position  will  be  equal  in  length 
to  the  assumed  portion  of  the  line  itself,  and  the  angle  which  this 
vertical  projection  of  the  line  in  this  new  position  makes  with  G-L 
will  indicate  the  angle  which  the  line  itself  in  true  position  makes 
with  H. 

109.  Problem  47.  Given  the  straight  line  [M  =  —  4,  5,  1 ;  N  —  4:^ 
2,  6] ;  required  to  find  the  true  distance  between  the  points  M  and  iV, 
and  to  determine  the  angle  which  the  line  makes  with  V. 

Analysis.  Revolve  the  line  about  an  axis  through  M  per- 
pendicular to  V  until  the  line  is  parallel  to  H.  The  horizontal 
projection  of  any  portion  of  the  line  in  this  new  position  will 
be  equal  in  length  to  the  assumed  portion  of  the  line  itself,  and 
the  angle  which  this  horizontal  projection  makes  with  G-L  will 
indicate  the  angle  which  the  line  itself  in  true  position  makes 
with  V. 

110.  Problem  48.  Draw  the  two  projections  of  a  straight  line  5 
units  long,  passing  through  the  point  M  =  2,  —  4,  —  1,  making  an 
angle  of  ZO  degrees  with  i/,  and  in  such  a  position  that  its  horizontal 
projection  makes  an  angle  of  45  degrees  with  G-L, 

Analysis.  First  draw  the  projections  of  the  line  when  it  passes 
through  the  given  point,  is  parallel  to  F,  and  makes  the  re- 
quired angle  with  H.  Then  revolve  the  line  about  an  axis 
through  the  given  point  perpendicular  to  H  until  the  horizontal 
projection  of  the  line  makes  the  required  angle  with  G-I^.  The 
two  projections  of  the  line  in  this  last  position  will  be  the  required 
projections. 


44 


DESCEIPTIVE  GEOMETRY 


Construction.  See  Fig.  54.  The  projections  of  the  point  M  are 
m,,  m'.  The  projections  of  the  line,  when  the  line  is  parallel  to  V 
and  makes  an  angle  of  30  degrees  with  i/,  are  m-nfj  and  m'-n'\ 
where  rrif-n,,  is  parallel  to  G-L  and  where  m'-n"  is  5  units  long 

and  makes   an  angle  of  30  degrees 
with  G-L. 

When  the  line  is  revolved  to  its 
final  position,  ?i,,  will  move  in  an  arc 
of  a  circle  to  n,,  and  n'^  will  move  in 
a  straight  line  parallel  to  G-L.,  to 
7i'.    The  required  projections  are  then 

771, 


n,  and  m' 


The  line  just  located  is  in  the 
third  quadrant  and  runs  from  the 
given  point  downward  to  the  left 
and  toward  V, 

Consider  other  positions  which  the  line  may  occupy  and  yet 
satisfy  the  requirements  of  the  problem. 

111.  Problem  49.    Construct  Problem  48  when  Jlif  =  —  2,  —  5,  1. 

112.  Problem  50.    Construct  Problem  48  when  M=  2,  5,  1. 

113.  Problem  51.    Construct  Problem  48  when  1/=  2,  2,  —  6. 

114.  Problem  52.  Draw  the  two  projections 
of  a  straight  line  4  units  long,  passing  through 
the  point  Jf  =  1,  —  1,  —  3,  7naking  an  angle  of 
45  degrees  with  F,  and  in  such  a  position 
that  its  vertical  projection  makes  an  angle  of 
60  degrees  with  G-L. 

Analysis.  First  draw  the  projections  of  the 
line  when  it  passes  through  the  given  point, 
is  parallel  to  H.,  and  makes  the  required 
angle  with  V.  Then  revolve  the  line  about 
an  axis  through  the  given  point  perpendicu- 
lar to  F,  until  the  vertical  projection  of  the 
line  makes  the  required  angle  with  G-L.  The  two  projections  of 
the  line  in  this  last  position  will  be  the  required  projections. 

Construction.    See  Fig.  55.    The  projections  of  the  given  point 
M  are  m,,  m\    The  projections  of  the  line,  when  the  line  is  parallel 


Fig.  55 


RELATING  TO  POINT,   LIXE,   AND  PLANE  45 

to  H  and  makes  an  angle  of  45  degrees  with  F,  are  m-n,^  and 
m'-n"^  where  m,-n,j  is  4  units  long  and  makes  an  angle  of 
45  degrees  with  G-L^  and  where  mJ-n"  is  parallel  to  G-L, 

When  the  line  is  revolved  to  its  final  position,  n"  will  move  in  the 
arc  of  a  circle  to  n'^  and  n,,  will  move  in  a  straight  line  parallel  to 
G-L,  to  Uj.    The  required  projections  are  then  m-n,  and  m'-n'. 

The  line  just  located  is  in  the  third  quadrant  and  runs  from  the 
given  point  upward  to  the  left  and  away  from  V. 

Consider  other  positions  which  the  line  may  occupy  and  yet 
satisfy  the  requirements  of  the  problem. 

115.  Problem  53.    Construct  Problem  52  when  ilf  =  —  3,  6,  —  2. 

116.  Problem  54.    Construct  Problem  52  when  M=  4,  6,  4. 

117.  Problem  55.    Construct  Problem  52  wheyi  Jf  =  —  4,  —  6,  2. 

118.  Problem  56.  Draw  the  two  projections  of  a  straight  line  6 
units  long,  passing  through  a  point  M  =  0,  —3,  —1,  makiiig  an 
angle  o/  60  degrees  with  H,  and  lying  in  a  profile  plane. 

Analysis.  First  draw  the  projections  of  the  line  when  it  passes 
through  the  given  point,  is  parallel  to  F,  and  makes  the  required 
angle  with  H.  .Then  revolve  the  line  about  an  axis  through  the 
given  point  perpendicular  to  H  until  the  horizontal  projection  of 
the  line  makes  the  proper  angle  with  G-L  (see  Section  104). 

119.  Problem  57.  Given  a  rectangular  card  whose  dimensions 
are  6  units  by  4  units,  whose  surface  is  parallel  to  F,  whose  long 
edges  are  perpendicular  to  If,  and  whose  upper  right-hand  vertex  is 
a  point  B  =  2,  —  1,  —  1  ;  required  (1)  to  draw  the  two  projections 
of  the  card  in  the  given  position^  and  (2)  to  revolve  the  card  about  its 
right-hand  vertical  edge  as  an  axis,  through  angles  of  SO  degrees, 
45  degrees,  60  degrees,  and  90  degrees,  and  to  draw  the  corresjjond- 
ing  projectio7is. 

Construction.  See  Fig.  56.  The  projections  of  the  card  in  its 
first  position  are  a-b^  and  a'-b'-d'-e',  where  b^  and  b'  are  the  pro- 
jections of  B,  where  a'-b'  is  4  units  long,  and  where  a'-e'  is  6 
units  long. 

When  the  card  is  revolved  about  B-D  as  an  axis  through  an 
angle  of  30  degrees,  the  two  projections  of  the  card  are  a^^-b^  and 
a"-b'-d'-e",  where  the  angle  a^-b^-a^^  is  30  degrees,  and  where  a" 
and  e"  are  in  a  straight  line  through  a,,  perpendicular  to  G-L. 


46 


DESCRIPTIVE   GEOMETRY 


When  the  card  is  revolved  through  an  angle  of  45  degrees,  its 
two  projections  will  be  ctj^^-h^  and  a'"-h'-d'-eJ" 

The  projections  of  the  card  in  the  different  positions  are  easily 
determined  when  we  remember  that  horizontal  projections   will 

move  in  arcs  of  circles  and  that 
vertical  projections  will  move  in 
straight  lines  parallel  to  G-L  (see 
Section  98). 

120.  Problem  58.  Given  a  rect- 
angular card  situated  in  the  third 
quadrant^  ivhose  dimensions  are  6 
units  hy  4  units^  whose  surface  is 
parallel  to  H^  and  whose  long  edges 
are  perpendicular  to  V ;  required 
(1)  to  draw  the  projections  of  the 
card  in  the  given  position^  and  (2)  to 
revolve  the  card  about  its  left-hand 
edge  as  an  axis,  through  angles  of 
30  degrees,  45  degrees,  60  degrees, 
and  90  degrees,  and  to  draw  the 
corresponding  projections. 

121.  Problem  59.  Given  a  rect- 
angular card  situated  in  the  first 
quadrant,  whose  dimensions  are  6 

units  hy  4  units,  whose  surface  is  parallel  to  V,  and  whose  long  edges 
are  perpendicular  to  II;  required  (1)  to  draw  the  projections  of  the 
card  in  the  given  position,  and  {2)  to  revolve  the  card  about  one  of  its 
vertical  edges  as  an  axis,  through  angles  o/  30  degrees,  45  degrees,  60 
degrees,  and  90  degrees,  and  to  draw  the  corresponding  projections. 

122.  Problem  60.  Given  a  rectangular  card  whose  dimensions  are 
5  units  by  3  units,  whose  surface  is  parallel  to  II,  whose  long  edges  are 
parallel  to  V,  and  whose  left-hand  back  vertex  is  A  =  —  S,  —  4,  —  1 ; 
required  (1)  to  draw  the  projections  of  the  card  in  the  given  position  ; 
{2)  to  revolve  the  card  about  its  left-hand  edge  as  an  axis  until  the 
surface  of  the  card  is  inclined  30  degrees  to  H,  and  to  draw  the  corre- 
sponding projections  ;  and  (S)  to  revolve  the  card  in  its  last  position 
about  a  vertical  axis  through  J  until  the  horizontal  projections  of  the 


Fig.  50 


RELATING  TO  POINT,  LINE,  AND  PLANE 


47 


edges  of  the  card  are  inclined  45  degrees  to  G-L^  and  to  draw  the 
corresponding  projections. 

Construction.  See  Fig.  57.  The  projections  of  the  card  in  its 
first  position  are  a-h^-d-e,  and  a'-h'-d'-e'. 

In  taking  its  second  position  the  card  is  revolved  about  an 
axis  through  A  perpendicular  to  V.  The  surface  of  the  card  will 
remain  perpendicular  to  Fand  its  vertical  projection  will  take  the 
position  a'-h"  equal  to  a -ft',  and  making  an  angle  of  30  degrees 
with  a'-h'.  Its  horizontal  projection  will  be  a-bf-d^^-Cf,  where 
A  and  U  have  made  no  change  in  position  and  where  5,,  and  c?„ 
fall  in  a  straight  line  through  b" 
perpendicular  to  G-L. 

In  taking  its  third  position  the 
card  is  revolved  about  an  axis 
through  A  perpendicular  to  Jf. 
The  horizontal  projection  of  the  . 
card  will  change  in  position  but 
not  in  character. 

Upon  a,,  which  makes  no 
change  in  position,  construct  the  q 
rectangle  a-bj^j-d^j-Cj,  equal  in 
every  respect  to  a-b^-df-e,  but 
having  its  edges  inclined  45  de- 
grees to  G-L.  This  is  the  hori- 
zontal projection  of  the  card  in 
its  final  position. 

The  vertical  projection  of  A  will  remain  at  a'.  The  vertical  projec- 
tion of  B  must  be  in  a  straight  line  through  b"  parallel  to  G-L^  and 
also  in  a  straight  line  through  5,^,  perpendicular  to  G-L,  or  at  b'"' 

The  vertical  projection  of  L>  must  be  in  a  straight  line  through 
d"  parallel  to  G-L,  and  also  in  a  straight  line  through  d,,,  per- 
pendicular to  G-L,  or  at  d^^'. 

The  vertical  projection  of  ^  must  be  upon  a  straight  line 
through  e'  parallel  to  G-L,  and  also  upon  a  straight  line  through 
e,f  perpendicular  to  G-L,  or  at  e'^. 

The  vertical  projection  of  the  card  in  its  final  position  is 
a'-b'"-d'"-e". 


48  DESCRIPTIVE  GEOMETRY 

123.  Problem  61.  G-iven  a  rectangular  card  whose  dimensions  are 
6  units  hy  4  units,  whose  surface  is  parallel  to  F,  whose  long  edges  are 
parallel  to  H,  and  whose  right-hand  upper  vertex  is  jE"  =  4,  —  2,  —  1 ; 
required  (1)  to  draw  the  projectioyis  of  the  card  in  the  given  position; 
\2)  to  revolve  the  card  about  its  right-hand  vertical  edge  as  an  axis 
until  the  surface  of  the  card  is  inclined  30  degrees  to  F,  and  to  draw 
the  corresponding  projections  ;  and  (3)  to  revolve  the  card  in  its  last 
position  about  an  axis  through  E  perpendicular  to  V  until  the  vertical 
projections  of  the  edges  of  the  card  are  inclined  45  degrees  to  G-L, 
and  to  draw  the  corresponding  projections. 

124.  Problem  62.  Solve  Problem  60,  substituting  a  new  value  for  A, 
namely  A=  —  4,  1,  1,  locating  the  whole  work  in  the  first  quadrant. 

125.  Problem  63.  Given  an  hexagonal  card  whose  side  is  3  units, 
whose  surface  is  parallel  to  F,  two  of  whose  sides  are  parallel  to  H, 
and  whose  extreme  left-hand  vertex  is  A  =  —  4,  —  1,  —  4  ;  required  (1)  to 
draw  the  projections  of  the  card  iri  the  given  position ;  {2)  to  revolve  the 
card  about  a  vertical  axis  through  A  until  the  surface  of  the  card  makes 
an  angle  of  QO  degrees  with  F,  and  to  draw  the  corresponding  projec- 
tions ;  and  (3)  to  revolve  the  card  in  its  last  position  about  an  axis 
through  A  perpendicular  to  V  until  the  vertical  projections  of  the  sides 
which  were  parallel  to  H  in  the  last  position  shall  make  an  angle  of 
45  degrees  with  G—L,  and  to  draw  the  corresponding  projections. 

Construction.  See  Fig.  58.  The  projections  of  the  card  in  its 
first  position  are  a-b-d-e^-f-g^  and  a! -V -W -e^ -f -g\ 

In  taking  its  second  position  the  card  is  revolved  about  a  ver- 
tical axis  through  A.  The  surface  of  the  card  will  remain  perpen- 
dicular to  H  and  its  horizontal  projection  will  take  the  position 
«,-g,,  equal  to  a^-e^,  and  making  an  angle  of  60  degrees  with  a,-e^. 
Its  vertical  projection  will  be  a'-b"-d"-e"-f'^-g",  where  a'  has 
made  no  change  in  position,  where  b"  is  at  the  intersection  of  a 
straight  line  through  b'  parallel  to  G-L,  with  a  straight  line 
through  bj^  perpendicular  to  G-L,  where  d"  is  at  the  intersection 
of  a  straight  line  through  d^  parallel  to  G—L,  with  a  straight  line 
through  dj,  perpendicular  to  G-L,  etc. 

In  taking  its  third  position  the  card  is  revolved  about  an  axis 
through  A  perpendicular  to  F.  The  vertical  projection  will  change 
in  position  but  not  in  character. 


KELATING  TO  POINT,  LINE,   AND  PLANE 


49 


Upon  a\  which  makes  no  change  in  position,  construct  the 
polygon  a'-b"'-d'"-e"'-f"'-g'"  equal  in  every  respect  to  a'-b"-d"- 
e"-f"-g'\  but  having  the  sides  which  were  originally  parallel  to 
G-L  now  inclined  at  an  angle  of  45  degrees  to  G-L.  This  is  the 
vertical  projection  of  the  card  in  its  final  position. 

Thd  horizontal  projection  of  A  will  remain  at  a^. 

The  horizontal  projection  of  B  must  be  in  a  straight  line  through 
bff  parallel  to  G-L,  and 
also   in  a  straight  line 
through  b"'  perpendicu- 
lar to  G-L,  or  at  5,,,. 

The  horizontal  pro- 
jection of  I)  must  be  in 
a  straight  line  through 
dff  parallel  to  G-L,  and 
also  in  a  straight  line 
through  d"'  perpendicu- 
lar to  G-L,  or  at  c?,„. 

The  horizontal  pro- 
jection of  LJ  must  be  in 
a  straight  line  through 
Cff  parallel  to  G-L,  and 
also 

through  e 
lar  to  G-L,  or  at  e,,,. 

The  horizontal  projec- 
tion of  the  card  in  its 
final  position  is  a-b,,- 

126.  Problem  64.  Given  an  hexagonal  card  whose  side  is  3 
units,  whose  surface  is  parallel  to  H,  two  of  whose  sides  are  parallel 
to  V,  and  whose  extreme  left-hand  vertex  is  A  =  —  4,  —  4,  —  1  ; 
required  (1)  to  draw  the  projections  of  the  card  in  the  given  position; 
{2)  to  revolve  the  card  about  an  axis  through  A  perpendicular  to  V  until 
the  surface  of  the  card  makes  an  angle  of  60  degrees  with  H,  and  to 
draw  the  corresponding  projections ;  and  {3)  to  revolve  the  card  in 
its  last  position  about  an  axis  tlirough  A  perpendicular  to  H  until 


in  a  straight  line 
'"  perpendicu- 


50 


DESCRIPTIVE   GEOMETRY 


the  horizontal  inojections  of  the  sides  which  ivere  parallel  to  V  in  the 
last  position  shall  he  inclined  45  degrees  to  G-L^  and  to  draiv  the 
corresponding  projections. 

127.  Problem  65.  Solve  Problem  63,  substituting  a  new  value  for 
A,  namelg  A=—4,  1,  4,  locating  the  whole  work  in  the  first  quadrant. 

128.  Problem  66.  To  fir^d  the  points  in  ivhich  a  given  straight 
line  intersects  H  and  V. 

Case  1.   To  find  the  poiyit  in  which  the  line  intersects  If. 

Analysis.  Since  the  point  in  which  the  line  intersects  H  must 
be  both  in  ^  and  in  the  line, itself,  its  vertical  projection  must  be 
both  in  G-L  and  in  the  vertical  projection  of  the  line,  and  there- 
fore at  their  intersection.    The  horizontal  projection  of  the  required 


/nuH 


Fig.  59 


Fk;.  00 


point  must  be  both  in  a  straight  line  perpendicular  to  G-L  through 
the  vertical  projection  just  found  and  also  in  the  horizontal  pro- 
jection of  the  line,  and  therefore  at  their  intersection. 

Since  the  required  point  is  in  II,  it  must  be  coincident  with  its 
horizontal  projection  just  found. 

Construction.  See  Figs.  59  and  60.  Let  M-j\^  represent  the 
given  line. 

Produce  the  vertical  projection  m'-7i'  to  meet  G~L  at  a'.  The 
point  a'  is  by  analysis  the  vertical  projection  of  the  required  point. 
At  a'  draw  a  straight  line  perpendicular  to  G-L  to  meet  the 
horizontal  projection  of  the  line  at  a„  which  is  both  the  horizontal 


EELATING  TO  POINT,  LINE,  AND  PLANE  51 

projection  of  the  point  and  the  point  itself  in  which  M-N  inter- 
sects H. 

Check.  See  Fig.  60.  Project  M-N  upon  the  supplementary 
plane  U^  which  is  assumed  parallel  to  the  horizontal  projecting 
plane  of  M-N. 

According  to  Section  55  the  line  mUj^-nUji  is  the  supplementary 
projection  of  M-N.  Produce  mUjj-nUjj  to  meet  the  horizontal  trace 
U-u,  in  au,^  which  is  the  supplementary  projection  of  the  point  in 
which  M-N  intersects  //.  According  to  Section  55  the  points  a, 
and  au,  should  be  in  the  same  straight  line  perpendicular  to  U-u^. 

Case  2.    To  find  the  point  in  which  the  line  intersects  V. 

Analysis.  Since  the  point  in  which  the  line  intersects  V  must 
be  both  in  V  and  in  the  line  itself,  its  horizontal  projection  must 
be  both  in  G-L  and  in  the  horizontal  projection  of  the  line,  and 
therefore  at  their  intersection.  The  vertical  projection  of  the 
required  point  must  be  both  in  a  straight  line  perpendicular  to 
G-L  through  the  horizontal  projection  just  found,  and  also  in  the 
vertical  projection  of  the  line,  and  therefore  at  their  intersection. 

Since  the  required  point  is  in  F,  it  must  be  coincident  with  its 
vertical  projection  just  found. 

Construction.  See  Figs.  59  and  60.  Produce  the  horizontal  pro- 
jection of  the  line  to  meet  G-L  at  6,.  At  this  point  draw  a  straight 
line  perpendicular  to  G-L  to  meet  the  vertical  projection  of  the 
line  at  h\  which  is  the  required  point. 

Chech.  See  Fig.  60.  Project  M-N  upon  the  supplementary 
plane  IF,  which  is  assumed  parallel  to  the  vertical  projecting 
plane  of  M-N^  and  proceed  as  in  Case  1. 

129.  Problem  67.  Given  the  straight  line  [M  =  —  6^  —  6,  2 ; 
N  =6^  2,  —  6] ;  required  to  find  the  points  in  which  M-N  intersects 
H  and  V. 

130.  Problem  68.  Given  the  straight  line  [ilf  =  —  6,  —4,  2; 
A^  =  6,  2,  2] ;  required  to  find  the  point  in  which  M-N  intersects  V. 

131.  Problem  69.  Given  the  straight  line  [M  =  Q,  2,  —  6 ; 
i\^  =  —  6,  2,  3] ;  required  to  find  the  point  in  ivhich  M-N  intersects  H. 

132.  Problem  70.  Given  the  straight  line  [M  =  —  Q,  6,  —6; 
iVr  =  6,  —  6,  6] ;  required  to  find  the  points  in  which  M-N  intersects 
H  and  F. 


52  DESCEIPTIVE  GEOMETRY 

133.  Problem  71.  Given  the  straight  line  [il^f  =  0,  —6,  —2; 
iV'=  0,  —  2,  —  6] ;  required  to  find  the  points  in  which  M-N  inter- 
sects H  and  V. 

134.  Problem  72.  To  find  the  distance  between  two  given  points 
in  space. 

This  distance  will  be  measured  on  the  straight  line  joining  the 
given  points. 

Analysis  1.  If  the  straight  line  connecting  the  two  points  be 
revolved  about  its  horizontal  projection  as  an  axis  into  H^  the  line 
will  be  seen  in  its  true  length  (see  Sections  26  and  87). 

Analysis  2.  If  the  straight  line  connecting  the  two  points  be 
revolved  about  its  vertical  projection  as  an  axis  into  F,  the  line 
will  be  seen  in  its  true  length. 

Analysis  3.  If  the  straight  line  connecting  the  two  points  be 
revolved  about  the  horizontal  projecting  line  of  one  of  its  points 
as  an  axis  until  the  line  is  parallel  to  F,  the  vertical  projection  of 
the  line  in  this  revolved  position  will  be  equal  in  length  to  the 
line  itself  (see  Sections  26  and  104). 

Analysis  4-    If  the  straight  line  connecting  the  two  points  be 

revolved  about  the  vertical  projecting  line  of  one  of  its  points  as 

n    ^^  ^^^^  until  the  line  is  parallel  to  H^  the 

/^  /       horizontal  projection  of  the  line  in  this 

revolved  position  will  be  equal  in  length 

\.     /  to  the  line  itself. 

Construction  1.    See   Fig.  61.    Let  M 

and  N  represent  the   two  given  points. 

Then  m-n^  will  represent  the  horizontal 

projection  and  mJ-n'  will  represent  the 

V:  ^  vertical   projection   of   the   straight  line 

connecting  these  points.    Following  An- 

^^^'  ^^  alysis  1,  revolve  M-N  about  m-n,  as  an 

axis  into  H  (see  Section  87).    M  falls  at  w^  and  N  falls  at  w^^, 

and  nijj-njj  is  the  distance  sought. 

Construction  2,  See  Fig.  62.  Let  M  and  N  represent  the  given 
points.  Following  Analysis  4,  revolve  M-N  about  the  vertical 
projecting  line  of  M  until  M-N  is  parallel  to  H  (see  Section  105). 
M  will  remain  stationary,  but  N  will  move,  its  vertical  projection 


\ 


RELATING  TO  POINT,  LINE,  AND  PLANE 


53 


Mrr 

IKx 

1                                  I'^'r 

a 

! 

' 

k 

J 

n'f 

[ \ 

m' 

Fig.  62 


taking  the  position  n"  and  its  horizontal  projection  taking  the 
position  n,f.    The  distance  m^-nj^  is  the  distance  sought. 

Check,  The  result  obtained  by  use  of  any  one  of  the  methods 
here  suggested  may  be  checked  by  using  one  of  the  other  methods. 

135.  Problem  73.  Find  the  distance 
between  the  two  points  Jf  =  —  6,  —  6,  2 
and  A^  =  6,  2,-4  hy  Analysis  2. 

136.  Problem  74.  Find  the  distance 
between  the  two  points  M  =—  Q^  6,  2  and 
N—  4:^  —  4:^  —  4:  by  Analysis  3. 

137.  Problem  75.  Find  the  distance 
between  the  two  points  Jf  =  0,  —  6,  1  and 
A^  =  0,  2,  8  by  any  method^  and  check 
it  by  another. 

138.  Problem  76.    To  pass  a  plane  through  three  given  points. 
Analysis.   If  we  connect  any  two  of  the  given  points  by  a  straight 

line,  such  a  line  must  lie  in  the  required  plane  and  must  pierce  H 
and  V  in  the  corresponding  traces  of  the  plane  (see  Section  42). 
A  straight  line  connecting  either  one  of  the  points  just  used  with 

the  remaining  point  must 
also  lie  in  the  required  plane 
and  therefore  must  pierce  H 
and  V  in  the  corresponding 
traces  of  the  required  plane. 
A  straight  line  obtained 
by  joining  any  one  of  the 
three  given  points  with  any 
point  in  a  straight  line  con- 
necting the  other  two,  or 
by  drawing  through  any 
one  of  the  three  points  a 
straight  line  parallel  to  the 
straight  line  connecting  the 
other  two,  is  a  line  of  the  required  plane. 

Construction.  See  Fig.  63.  Let  Jf,  A,  and  0  represent  the  three 
given  points.  Connect  Jf  and  A"  by  a  straight  line  and  produce  it 
to  intersect  H  in  a,  and  V  in  b'.    The  point  a^  is  a  point  in  the 


Fig.  63 


54  DESCRIPTIVE  GEOMETRY 

horizontal  trace  of  the  required  plane,  and  the  point  h'  is  a  point 
in  the  vertical  trace.  Connect  N  and  the  remaining  point  0  by  a 
straight  line.  The  point  e,  in  which  N-0  intersects  H  is  another 
point  in  the  horizontal  trace,  and  the  point  d'  in  which  N-0  inter- 
sects V  is  another  point  in  the  vertical  trace.  S-s,  drawn  through 
a^  and  e,  is  the  horizontal  trace  of  the  required  plane,  and  S-s' 
drawn  through  h'  and  d^  is  the  vertical  trace. 

Check.    The  two  traces  should  intersect  in  G-L. 

139.  Problem  77.  Determine  the  traces  of  a  plane  containing  the 
straight  line  [Jf  =  —  4,  Q.,  ^]>  iV^=  2,  —  2,  1]  and  the  point  0  =  4, 
2,  -  3. 

140.  Problem  78.  Determine  the  traces  of  a  plarie  containing  the 
two  lines  [Jf  =  -  6,  -  6,  4;  N=  2,  2,  -  4]  and,  [0  =  -  2,  -  7,  4; 
P  =  6,  1,-4]. 

141.  Problem  79.  Determine  the  traces  of  a  plane  containing  the 
two  lines  [ilf  =  -  4,  1,  4;  i\^==  4,  1,  4]  and  [0  =  -  6,  4,  1 ;  P  =  Q, 
4,  1]. 

142.  Problem  80.  To  find  the  angle  between  two  intersecting 
straight  lines,  and  to  bisect  the  angle. 

Analysis  1.  Revolve  the  plane  of  the  angle  about  its  horizontal 
trace  or  its  vertical  trace  into  the  corresponding  plane  of  projec- 
tion. The  angle  between  the  two  lines  will  then  be  seen  in  its 
true  size  (see  Section  87),  and  may  be  bisected. 

Analysis  2.  Revolve  the  plane  of  the  angle  about  some  straight 
line  which  is  in  the  plane  and  parallel  to  one  of  the  planes  of  pro- 
jection, until  the  plane  is  parallel  to  that  plane  of  projection.  The 
projection  of  the  angle  upon  this  plane  of  projection  will  then 
be  equal  to  the  angle  itself  and  may  be  measured  and  bisected. 

Construction.  See  Fig.  64.  Let  Jf-iVand  N-0.,  intersecting  at 
iV,  represent  the  two  given  lines.  Following  Analysis  1,  produce 
M-N  to  meet  H  at  a,  and  produce  N-0  to  meet  H  at  b,.  The  line 
Sf-S-Sf  through  a,  and  6,  is  the  horizontal  trace  of  the  plane  of  the 
given  angle.  Revolve  this  plane  about  s^f-S-s^  as  an  axis  into  IT. 
The  points  a,  and  5,,  which  are  points  in  the  sides  of  the  angle  and 
also  in  the  axis,  will  remain  stationary.  The  vertex  N  will  fall  at 
njj  (see  Section  87).  The  angle  a-nj-b^  is  the  required  angle. 
Bisect  this  angle  by  the  line  nj^-d,  intersecting  s^-S-s,  at  d,. 


KELATING  TO  POINT,  LINE,  AND  PLANE 


55 


When  the  plane  of  the  angle  is  revolved  back  to  its  original 
position,  dj  will  remain  stationary  and  N  will  take  its  old  position. 
The  horizontal  projection  of  the  bisector  in  true  position  will  be  at 
n^-d,  and  its  vertical  projection  will  be  at  n'-d'. 

Check.  Revolve  the  plane  of  the  angle  about  its  vertical  trace 
into  the  vertical  plane  of  projection. 

143.  Problem  81.  Determine  the  angle  between  the  two  lines 
[M  =  -6,  4,-2;  JSr=0,  -4,  -6]  a7id  [iV^=  0,  -4,  -6; 
0  =  6,  2,  1]. 

144.  Problem  82.  Determine  the  migle  between  the  two  lines 
[Jf  =  -4,  5,  6;    A^=4,  1,  2]    and   ^Y-4,  1,  2 ;    0  =  4,  6,  2]. 


Fig.  64 


145.  Problem  83.  Determine  the  angle  between  the  two  lines 
[j/  =  _4,  _6,  -6;  .¥=4,  -2,  -2]  and  ^Y=4,  -2,  -2; 
0  =  4,  -2,  -5]. 

146.  Problem  84.  To  find  the  intersection  of  two  planes  when 
their  traces  are  Jcnow7i. 

Case  1.  When  the  horizontal  traces  of  the  two  planes  intersect 
within  the  limits  of  the  drawing  and  when  the  vertical  traces  of  the 
same  planes  also  intersect  within  the  limits  of  the  drawing. 

Analysis.  The  intersection  of  the  horizontal  traces  must  be  a 
point  common  to  both  planes  and  therefore  a  point  in  their  inter- 
section. For  the  same  reason  the  intersection  of  the  vertical  traces 
must  be  another  point  in  the  required  intersection. 


56 


DESCRIPTIVE  GEOMETRY 


Since  the  intersection  of  two  planes  is  a  straight  line,  the 
straight  line  joining  these  two  points  must  be  the  required  line 
of  intersection. 

Co7istructio7i.  See  Fig.  65.  Let  iS'and  T  represent  the  two  given 
planes.    The  horizontal  traces  intersect  at  a^  vertically  projected  at  a'. 

The  vertical  traces  intersect  at 
b'  horizontally  projected  at  5,. 
According  to  analysis  A-B  is 
the  required  intersection. 
Cheek.   Connect  any  point  in 
jj  the  line  of  intersection  with  any 
point  in  the  vertical  trace  of  one 
of  the  planes,  and  note  whether 
this  line  intersects  IT  in  the  hori- 
zontal trace  of  the  same  plane. 
Apply  the  same  test  with  ref- 
^iG.  65  erence  to  the  other  plane. 

Case  2.  When  either  the  two  horizontal  traces  or  the  two  vertical 
traces  do  not  intersect  within  the  limits  of  the  drawing. 

Analysis.  Pass  a  series  of 
auxiliary  planes  parallel  to  the 
plane  of  projection  on  which 
the  traces  intersect.  These 
planes  will  cut  from  the  two 
given  planes  straight  lines 
which  will  be  parallel  to  tliis 
plane  of  projection  and  which 
will  intersect  in  points  com- 
mon to  both  of  the  given 
planes  and  therefore  in  their 
line  of  intersection. 

Construction.  See  Fig.  66. 
Let  S  and  T  represent  the 
given  planes  whose  horizontal 
traces  do  not  intersect  within  the  limits  of  the  drawing.  The  ver- 
tical traces  intersect  at  a'  horizontally  projected  at  a,,  locating  one 
point  in  the   required   intersection.     Pass   an   auxiliary  plane   U 


Fig.  m 


RELATmO  TO  POINT,  LINE,   AND  PLANE 


57 


parallel  to  V.  The  plane  U  cuts  the  plane  iS  in  a  straight  line 
B-U  parallel  to  S-s'  and  vertically  projected  in  b'-e'  parallel  to 
S-s'  (see  Sections  30  and  41).  The  plane  U  cuts  the  plane  T  in  the 
line  D-E  vertically  projected  in  d'-e'.  The  point  e'  in  which  the 
two  vertical  projections  intersect  must  be  the  vertical  projection  of 
the  point  in  which  the  lines  B-E  and  D-U  intersect,  and  is,  accord- 
ing to  analysis,  the  vertical  projection  of  a  point  in  the  required 
intersection.  Since  the  point  JS  is  in  the  plane  £/,  its  horizontal 
projection  e^  must  be  upon  U-u^,  and  therefore  at  the  intersection 
of  U-Uf  and  a  straight  line  through  e'  perpendicular  to  G-L. 

Pass  another  auxiliary  plane  W  parallel  to  F,  and  in  the  same 
manner  obtain  K^  another  point  in  the  required  intersection. 

The  line  A-E-K^  whose  liorizontal  projection  is  a-e^-k^  and 
whose  vertical  projection  is  a'-e'-k'^  is  the  required  line. 

Check.  The  straight  line  E-K  thus  determined  by  the  two  points 
E  and  K  should  pass  through  A. 

147.  Problem  85.  Find  the  intersection  of  the  two  planes  aS  =  —  2, 
330°,  300°,  and  T=  0°,  -  3,  x. 

148.  Problem  86.  Find  the  intersec- 
tion of  the  two  planes  /S^  =  — 2,  60°, 
30°,  and  T=  0°,  oc,  2. 

149.  Problem  87.  Find  the  intersec- 
tion of  the  two  planes  ^S  =  —  6,  30°,  90°, 
and  T=6,  135°,  90°. 

150.  Problem  88.  Find  the  intersec- 
tion of  the  two  planes  S  =  0°,  6,  2,  and 
T=0%  2,  6. 

151.  Problem  89.  To  find  the  point 
in  which  a  given  straight  line  intersects 
a  given  plane. 

Analysis.  The  given  line  must  intersect  the  given  plane  in  the 
line  in  which  any  auxiliary  plane  containing  the  given  line  inter- 
sects the  given  plane.  The  point  in  which  the  given  line  crosses 
this  line  of  intersection  must  be  the  required  point. 

Case  1.    Whe7i  the  given  plane  is  located  hy  its  traces. 

Construction.  See  Fig.  67.  Let  M-N  represent  the  given  line 
and  let  S  represent  the  given  plane. 


58 


DESCRIPTIVE  GEOMETRY 


For  ease  of  construction  let  us  take  the  horizontal  projecting 

plane  of  M-N  as  the  auxiliary  plane.    This  plane  T  intersects  S  in 

A-B  (see  Section  146),  which  is  cut  by  M—No^i  the  required  point  D. 

Check,    Use  the  vortical  projecting  plane  of  M—N. 

Case  2.     When   the  given  plane   is   located  hy  two   intersecting 

straight  lines. 

Construction.    See  Fig.  68.    Let  iT/-iYand  0-F,  intersecting  at  X, 

locate  the  given  plane,  and  let  Q-E  represent  the  intersecting  line. 

By  Case  1  above,  If-iV  intersects  the  horizontal  projecting  plane 

of  Q-E  at  A.    0-P  intersects  the  same  plane  at  B.    Therefore  A-B 

is  the  intersection  of  the 
horizontal  projecting 
plane  of  Q-E  and  the 
plane  of  the  two  given 
lines  M-N  and  0-P. 
Q-E  crosses  A-B  at  D, 
which  by  analysis  is 
the  required  point  of 
intersection. 

Chech.  Use  the  ver- 
tical projecting  plane  of 
Q-E. 

152.  Problem  90.  Find 
the  point  in  which  the  j^Zawe  determined  hy  the  two  liyies  \_M  =  —  6, 
6,  6  ;  .¥•=  0,  2,  1]  and  [il^=  -  6,  6,  6 ;  0=2,  8,  4]  is  intersected 
hy  the  line  [(^  =  -  4,  2,  2 ;  i^  =  2,  6,  8]. 

153.  Problem  91.  Find  the  point  in  which  the  plane  determined  hy 
the  two  lines  [J/=  -  4,  -  2,  -  8  ;  7V^  =  -  4,  -  8,  -  2]  and  [M  =  -  4, 

—  2,  —  8 ;  0  =  4,  —  7,  —  3]  is  intersected  hy  the  line  [(^  =  —  6,  —  6, 

-  2  ;  i^  =  6,  -  6,  -  8]. 

154.  Problem  92.  To  draw  through  a  given  point  a  straight  line 
perpendicular  to  a  given  plane,  and  to  determine  the  distance  of  the 
point  from  the  playie. 

Analysis.  By  Section  44  we  know  that  the  horizontal  projection 
of  the  required  line  must  be  perpendicular  to  the  horizontal  trace 
of  the  given  plane,  and  that  the  vertical  projection  of  the  required 
line  must  be  perpendicular  to  the  vertical  trace. 


KELATIKG  T(J  POINT,  LINE,  AND  PLANE 


59 


The  distance  of  a  point  from  a  plane  is  measured  on  a  straight 
line  drawn  from  the  point  perpendicular  to  the  plane.  If  then  we 
find  the  point  in  which  such  a  line  intersects  the  given  plane,  the 
distance  from  the  given  point  to  this  point  of  intersection  will  be 
the  required  distance. 

Construction.  See  Fig.  69.  Let  S  represent  the  given  plane  and 
let  0  represent  the  given  point.  Through  o,  draw  o^-d^  perpendicular 
to  S-Sf  and  through  o'  draw  o'-d'  perpendicular  to  >S-s'.  0-D  is 
the  required  line. 

By  Problem  89  find  the  point  D  in  which  the  line  0-D  intersects 
S.  The  distance  0-D^  which  is  found  by  revolving  the  horizontal 
projecting  plane  of  0-D  about  o^-d,  into  H  (see  Problem  72),  is 
the  required  distance. 

Chech.  To  tes^;  the  perpendicu- 
larity of  the  line,  draw  through  D 
some  line  of  the  plane  and  note 
whether  this  line  is  perpendicular  to 
0-D.  To  test  the  distance  of  the 
point  from  the  plane,  revolve  the 
vertical  projecting  plane  of  0-D 
about  o'-d'  into  V. 

155.  Problem  93.  Fiyid  the  dis- 
tance of  the  point  J/=  —  2,  — 5,  — 8 
from  the  plane  S  =  0°,  gc,  2. 

156.  Problem  94.  Find  the  dis- 
tance of  the  point  Jif  =  —  2,  —  6,  —  6 
from  the  plane  S  =  0°,  2,  gc. 

157.  Problem  95.  Find  the  distance  of  the  point  Jf  =  0,  —  6,  —  6 
from  the  plane  ^  =  -  4,  300°,  270°. 

158.  Problem  96.  Find  the  distance  of  the  point  M  =  0,-6,-6 
from  the  plane  S  =  -  4:,  270°,  330°. 

159.  Problem  97.  Find  the  distance  of  the  point  Jf  =  0,  —  8,  6 
from  the  plane  S=  0,  210°,  60°. 

160.  Problem  98.  Find  the  distance  of  the  point  JIf  =  0,  —  6,  —  6 
from  the  playie  S  =  0°,  —3,-6. 

161.  Problem  99.  To  project  a  given  straight  line  upon  a  given 
plane. 


Fig.  69 


60 


DESCEIPTIVE  GEOMETRY 


— L 


Principle.  The  projection  of  a  straight  line  upon  a  plane  is  the 
line  in  which  a  plane  drawn  through  the  given  line  perpendicular 
to  the  given  plane  intersects  the  given  plane  (see  Section  23). 

Analysis  1.  If  through  any  point  of  the  given  line  we  draw  a 
straight  line  perpendicular  to  the  given  plane,  the  plane  of  the 
given  line  and  this  auxiliary  line  will  be  the  projecting  plane  of 
the  given  line,  and  will  intersect  the  given  plane  in  the  required 
projection. 

Analysis  2.  Since  the  projection  of  a  straight  line  upon  a  plane 
is  a  straight  line  and  contains  the  projections  of  all  the  points  of 

the  given  line,  the  required 
projection  is  the  straight  line 
determined  by  the  projections 
of  any  two  points  of  the 
given  line. 

Construction.  See  Fig.  70. 
Let  M-N  represent  the  given 
line  and  let  S  represent  the 
given  plane.  Following  An- 
alysis 2,  project  upon  the 
given  plane  any  two  points 
of  the  given  line,  as  M  and 
iV.  The  projection  of  M 
upon  S  is  MS^  horizontally 
projected  at  mSj  and  verti- 
cally projected  at  ms'  (see  Sections  5  and  151). 

The  projection  of  N  upon  S  is  NS^  horizontally  projected  at  ns, 
and  vertically  projected  at  ns' .  Therefore  ms-ns,  is  the  horizontal 
projection  and  ms'-ns'  is  the  vertical  projection  of  the  required 
projection. 

Check.  Assume  a  third  point  0  upon  the  given  line.  Project 
this  point  upon  S,  and  note  if  this  projection  falls  upon  the 
projection  already  found. 

162.  Problem  100.  Project  the  line  [1/=  -  2,  -  3,  -  8 ;  iV^=  3, 
—  9,—  4]  upon  the  plane  S  =  —  8,  315°,  SSO° ,  followiny  Analysis  1. 

163.  Problem  101.  Project  the  line  [Jf  =  -  6,  6,  8 ;  i\^=  6,  -  3, 
6]  upon  the  plane  >S  =  0°,  —  3,  6. 


J^i-^ 


Fig.  70 


EELATi:N^a  TO  POINT,  LINE,  AND  PLANE 


61 


164.  Problem  102.   Project  the  line  [Jf  =  —  4,  —  4,  —  4  ;  N=0, 
—  8,-6]  upo7i  the  plane  /S^  =  —  6,  315°,  ^S0° ,  following  Analysis  2, 

165.  Problem  103.    To  pass  a  plane  through  a  given  point  and 
perpendicular  to  a  given  straight  line. 

Analysis.  From  Section  44  we  know  that  the  horizontal  and 
vertical  traces  of  the  required  plane  will  be  perpendicular  respec- 
tively to  the  horizontal  and  vertical  projections  of  the  given  line. 
We  know,  then,  the  direction  of  each  of  the  required  traces.  If  a 
straight  line  be  drawn  through  the  given  point  parallel  to  either 
of  these  traces,  it  must  be  a  line  of  the  required  plane,  and,  unless 
parallel  to  G-L^  will  intersect  one  of  the  planes  of  projection  in 
a  point  of  the  corresponding  trace  of  the  required  plane.  A 
straight  line  through  the  point  thus  found  and  perpendicular 
to  the  corresponding  projection  of 
the  given  line  will  be  one  of  the 
required  traces. 

The  other  required  trace  will  pass 
through  the  point  in  which  the  trace 
just  found  intersects  G-L^  and  will 
be  perpendicular  to  the  remaining 
projection  of  the  given  line. 

Construction.  See  Fig.  71.  Let 
M-N  represent  the  given  line,  and 
let  0  represent  the  given  point. 
Through  0  draw  0-A  parallel  to  the 
horizontal  trace  of  the  required  plane. 
Since  this  horizontal  trace  is  to  be  perpendicular  to  m,- 


FiG.  71 


the 


horizontal  projection  of  0—A  will  be  perpendicular  to  m—n^. 
Since  0-A  is  parallel  to  a  line  in  If,  its  vertical  projection  will 
be  parallel  to  G-L. 

The  point  a'  in  which  0-A  intersects  F  is  a  point  in  the  vertical 
trace  of  the  required  plane,  and  S-s^  drawn  through  a'  perpen- 
dicular to  m'-n'  is  the  required  vertical  trace.  S-s^  drawn  through 
>S^  perpendicular  to  m-n^  is  the  required  horizontal  trace. 

Cheek.  Draw  a  line  through  0  parallel  to  the  required  vertical 
trace,  and  note  whether  such  a  line  intersects  If  in  the  horizontal 
trace  already  found. 


62 


DESCKIPTIVE  GEOMETRY 


166.  Problem  104.  Draiv  a  plane  through  the  pohit  0  =  —  2,  4,  2 
and  perpendicular  to  the  line  [Jf  =  —  6,  —  6,  —  6  ;  A^=  6,  —  3,  —  1]. 

167.  Problem  105.  Draw  a  plane  through  the  point  0  =:  2,  —6,  —  2 
and  perpendicular  to  the  line  [M=—  2,  1,  6  ;  N  =  —  2,  6,  1]. 

168.  Problem  106.  Dratv  a  plane  through  the  point  0=— 4,  9,  4 
and  perpendicular  to  the  litie  [J/=—  6,  —  4,  4  ;  N=  6,  3,  —  3]. 

169.  Problem  107.  To  pass  a  plane  through  a  given  poiyit  and 
parallel  to  two  given  straight  lines. 

Analysis,  If  through  the  given  point  we  draw  two  straight 
lines  parallel  respectively  to  the  two  given  lines,  the  plane  of 
these  two  lines  will  be  parallel  to  the  two  given  lines,  since  a  plane 
is  parallel  to  a  straight  line  when  it  contains  a  straight  line  parallel 
to  that  line. 

Construction.  See  Fig.  72.  Let  ilf-i\^  and  0-P  represent  the 
two  given  lines,  and  let  Q  represent  the  given  point.     Through  Q 

draw    A-B   parallel    to 
also  through    Q 
C-D  parallel   to 
A-B  intersects 
h^  and  intersects 
V  at  a!.      C-D   inter- 
sects H  at  c,  and  inter- 
sects Fat  c7'.    Therefore 
iS-Sp    which    is    drawn 
through  h^  and  c,,  is  the 
horizontal  trace  of  the 
required  plane,  and  S-s\ 
which  is  drawn  through 
a'  and  d\  is  the  vertical  trace  of  the  required  plane. 

Check.  By  Section  43  assume  any  point  in  the  plane  now 
determined.  Through  this  point  draw  two  straight  lines,  one 
parallel  to  one  of  the  given  lines  and  the  other  parallel  to  the 
other.  The  points  in  which  these  two  lines  intersect  H  and  V 
should  lie  in  the  horizontal  and  vertical  traces  already  located. 

170.  Problem  108.  Pass  a  plane  through  the  point  (>  =  —  4,  —  4,  —  2 
and  parallel  to  the  two  lines  \_M  =  0,  2,  6  ;  ^=  8,  —  3,  1]  and  [0  =  4, 
4,1;  P  =  4,  -2,6]. 


RELATING  TO  POINT,   LINE,  AND  PLANE 


63 


171.  Problem  109.  Pass  a  plane  through  the  point  Q——  2,  —  6,  3 
and  parallel  to  the  two  lines  [M=  —  4,  5,  —  3  ;  i\^=  4,  —  3,  6]  and 
[0=-2,  -1,5;  F  =  6,  -1,5]. 

172.  Problem  110.  To  pass  a  plane  through  a  given  point  and 
parallel  to  a  given  plane. 

Analysis.  The  traces  of  the  required  plane  will  be  parallel  to 
the  corresponding  traces  of  the  given  plane,  and  will  be  fully 
known  when  one  point  in  each  trace  is  determined.  A  straight 
line  through  the  given  point  and  parallel  to  either  trace  of  the 
given  plane  will  be  a  line  of 
the  required  plane  and  will  in- 
tersect ^  or  F  in  the  corre- 
sponding trace  of  the  required 
plane. 

Construction,  In  Fig.  73  let 
S  represent  the  given  plane  and 
let  Jf  represent  the  given  point. 

Through  31  draw  M-A  par- 
allel to  S-s'  and  produce  it  to 
meet  H  at  a^.  The  point  a,  is 
in  the  horizontal  trace  of  the 
required  plane. 

Through  31  draw  3I-B  parallel  to  S-s,  and  produce  it  to  meet 
V  at  6'.  The  point  b'  is  in  the  vertical  trace  of  the  required 
plane. 

The  two  required  traces,  T-t^  and  T-t',  the  former  parallel  to 
S—Sf  and  the  latter  parallel  to  S-s',  are  now  located. 

Check.    The  two  traces  now  determined  should  meet  on  G-L. 

173.  Problem  111.  Pass  a  plane  through  the  point  3£=— 6,— 4:,— 6 
and  parallel  to  the  plane  S  =—  6,  40°,  60°. 

174.  Problem  112.  Pass  a  plane  through  the  point  31=0.,  3,  6 
and  parallel  to  the  plane  S  =  0°,  5,  3. 

175.  Problem  113.  To  pass  a  plane  through  a  given  straight  line 
and  parallel  to  another  given  straight  line. 

Analysis.  Through  any  point  of  the  first  line  draw  a  straight 
line  parallel  to  the  second  line.  The  plane  of  the  first  line  and 
the  auxiliary  line  will  be  the  required  plane. 


Fig.  73 


64 


DESCEIPTIVE  GEOMETKY 


Construction.  See  Fig.  74.  Let  y¥-i\^  represent  the  line  through 
which  the  plane  is  to  be  passed,  and  let  0-P  represent  the  line  to 
which  the  plane  is  to  be  parallel. 

Through  any  point  of  M-N,  as  M,  draw  M-Q  parallel  to  0-P 


and  produce  it  to  intersect  H  in  c^  and  to  intersect  V  in  d'.    Find 


the  point  a,  in  which  M-N 
intersects  H-,  also  find  the 
point  h'  in  which  M-N  in- 
tersects V.  Through  c^  and 
a  J  draw  S-s,^  and  through  h' 
and  d'  draw  S-s'.  These  are 
the  traces  of  the  required 
plane  and  should  cross 
G-L  at  the  same  point. 

Check.  Through  some 
other  point  of  M-N  draw 
another  straight  line  paral- 
1  e  1  to  0-P  and  note 
the  traces  now  located. 
6,2,6;.Y=4, 


Fig.  74 

whether  this  line  intersects  H  and  V  in 

176.  Problem  114.   Pass  a  plane  through  [M  = 
—  5,  1]  and  parallel  ^o  [0  =1,  —  1,  6  ;  P  =  1,  5,  2]. 

177.  Problem  115.  Pass  a  plane  through  [if  =  5,  —6,  —3; 
iV  =  -4,  1,  8]  and  parallel  ^(?  [0  =  -  5,  3,  5  ;  P  =  2,  3,  5]. 

178.  Problem  116.  Pass  a  plane  through  [7l[f  =  —  6,  4,  —  6 ;  N 
=  5,  4,  -  6]  and  parallel  to  [0  =  2,  2,  6  ;  P  =  2,  -  6,  1]. 

179.  Problem  117.  To  find  the  shortest  distance  from  a  given 
point  to  a  given  straight  line. 

The  required  distance  must  be  measured  upon  a  straight  line 
drawn  from  the  given  point'perpendicular  to  the  given  line. 

Analysis  1.  If  we  pass  a  plane  through  the  given  point  and  the 
given  line,  and  revolve  this  plane  about  one  of  its  traces  into  the 
corresponding  plane  of  projection,  the  line  and  the  point  will  be 
shown  in  their  true  relation.  From  the  revolved  position  of  the 
point  draw  a  straight  line  perpendicular  to  the  revolved  position 
of  the  line,  and  upon  this  line  measure  the  required  distance. 

Analysis  2.  Draw  through  the  given  point  a  plane  perpendicu- 
lar to  the  given  line  and  find  the  point  in  which  this  plane  is 


RELATING  TO  POINT,  LINE,  AND  PLANE 


65 


intersected  by  the  given  line.  A  straight  line  from  this  point  of 
intersection  to  the  given  point  will  be  perpendicular  to  the  given 
line  and  will  therefore  be  the  line  upon  which  the  required  distance 
may  be  measured. 

Construction.  See  Fig.  75.  Let  M-N  represent  the  given  line 
and  let  0  represent  the  given  point. 

Following  Analysis  1,  draw  the  line  0-P  through  0  parallel  to 
M-N  and  produce  it  to  intersect  if  at  pj.  Find  <x,,  the  point  in 
which  M-N  intersects  H.  S-s,^-  drawn  through  p^  and  a,,  is  the 
horizontal  trace  of  a  plane 
containing  M-N  and   0. 

Revolve  this  plane  about 
S-8,  as  an  axis,  into  H.  0  will 
fall  at  0^,  M  will  fall  at  m^ 
and  <x,  will  remain  stationary. 
Since  A  and  Jtf  are  points  of  the 
line  Jf-iV,  <x,-m^is  the  revolved 
position  of  the  given  line. 

Through  Ojj  draw  Ojj-hjj 
perpendicular  to  a^-nijj.  The 
distance  Oj^-hjj  is  the  required 
distance. 

Check.  Revolve  the  plane 
of  the  point  and  the  line 
about  its  vertical  trace  into  F,  or  solve  according  to  Analysis  2. 

180.  Problem  118.  Find  the  distance  from  the  point  0  =  2,  4,  6 
to  the  line  [^  =  -  4,  -  6,  4  ;  A^  =  4,  -  6,  -  4]. 

181.  Problem  119.  Find  the  distance  from  the  point  0  =  4,  6,  2 
to  the  line  [M  =  ~  2,  2,  6  ;  A^=  -  2,  -  6,  -  5]. 

182.  Problem  120.  To  draw  through  a  given  point  a  straight  line 
perpendicular  to  a  given  straight  line. 

Analysis.  This  problem  is  embodied  in  Problem  117,  since  in  find- 
ing the  shortest  distance  from  a  point  to  a  straight  line,  a  straight  line 
is  drawn  from  the  given  point  perpendicular  to  the  given  line. 

If  in  the  solution  of  Problem  117,  Analysis  2  is  followed,  the 
projections  of  the  line  required  in  Problem  120  will  be  found 
directly. 


Fig.  76 


66  DESCRIPTIVE  GEOMETRY 

If  in  the  solution  of  Problem  117,  Analysis  1  is  followed,  a 
counter  revolution  will  be  necessary  in  order  to  determine  the 
projections  of  the  line  required  in  Problem  120. 

Construction.  Making  use  of  the  results  obtained  in  Fig.  75, 
revolve  the  plane  S  back  to  its  original  position.  0  will  take  the 
position  ((?,,  o')  which  it  originally  occupied,  and  B^  which  is  a 
point  on  M-N^  will  move  back  in  a  plane  perpendicular  to  S-s, 
until  its  horizontal  projection  takes  the  position  hj  and  its  vertical 
projection  takes  the  position  h'  in  the  straight  line  througli  h^ 
perpendicular  to   G-L. 

The  lines  Oj-h,  and  o'-h'  represent  respectively  the  horizontal 
and  vertical  projections  of  the  required  line  through  0  perpendic- 
ular to  M-N. 

Chech,    Follow  Analysis  2,  Problem  117. 

183.  Problem  121.  Draw  through  the  point  0  =  0,  4,  —  2  a  straight 
line  perpendicular  to  the  line  [1^1=  0,  2,  6 ;  A^=  0,  —  5,  —  2]. 

184.  Problem  122.  Draw  through  the  point  0  =  4,  —  2,  5  a  straight 
line  perpendicular  to  the  line  [M=  —  4,  2,  — 5;  A^=  —  4,  —  3,6]. 

185.  Problem  123.  To  find  the  angle  which  a  given  straight  line 
makes  with  a  given  plane. 

The  angle  which  a  straight  line  makes  with  a  plane  is  the  angle 
which  the  line  makes  with  its  projection  on  that  plane. 

Analysis  1.  By  Problem  99  project  the  given  line  upon  the 
given  plane,  and  by  Problem  80  measure  the  angle  between  the 
line  itself  and  its  projection. 

Analysis  2.  If  through  any  point  of  the  given  line  a  straight 
line  be  drawn  perpendicular  to  the  given  plane,  it  will  inter- 
sect the  given  plane  in  one  point  of  the  projection  of  the  given 
line  upon  this  plane.  The  line  itself  will  intersect  the  plane 
in  another  point  of  this  projection.  A  straight  line  through  the 
two  points  just  found  will  be  the  projection  of  the  line  upon  the 
plane. 

The  line  itself,  the  projection  of  the  line,  and  the  straight  line 
drawn  perpendicular  to  the  plane,  together,  form  a  right-angled 
triangle.  In  this  triangle  the  line  itself  is  the  hypotenuse,  the  pro- 
jection of  the  line  is  the  base,  and  the  straight  line  perpendicular 
to  the  plane  is  the  altitude. 


RELATING  TO  POINT,  LINK,  AND  PLANE 


67 


The  line  a,-5, 


The  oblique  angle  at  the  base  of  the  triangle  is  the  required 
angle,  and  since  the  triangle  is  right-angled  the  angle  at  the 
vertex  must  be  the  complement  of  the  required  angle. 

Construction.  See  Fig.  76.  Let  S  represent  the  given  plane  and 
let  M-N  represent  the  given  line. 

Following  Analysis  2,  through  some  point  of  M-N,  as  M, 
draw  M-B  perpendicular  to  S  and  produce  it  to  intersect  j^  at  h,. 
Find  the  point  a,  in  which  M-N  intersects  H. 
is  the  horizontal  trace  of  the 
plane  of  the  two  lines  M-N 
and  M-B. 

Revolve  this  plane  about 
a-h,  as  an  axis,  into  H.  The 
points  a,  and  h^  remain  station- 
ary and  M  revolves  to  rrij^. 
The  angle  aj-mjj-h,  is  the  true  q 
size  of  the  angle  between  the 
given  line  and  the  line  perpen- 
dicular to  the  plane,  and  is, 
by  analysis,  the  complement  of 
the  required  angle. 

Draw  nijj-ejj  perpendicular  to  mjj-hj.  The  angle  a^-mjj-ejj.^  which 
by  construction  is  the  complement  of  the  angle  af-m^j-b^,  is  the 
required  angle. 

Check.  Solve  by  Analysis  1,  or  assume  some  point  other  than 
Jf,  in  M-N,  as  the  point  through  which  to  draw  the  line  perpen- 
dicular to  the  plane. 

186.  Problem  124.   Find  the  angle  between  the  line  [Jf  =  —  6,  3,  5 
N=  5,  -4,  2]  and  the  plane  S=  0°,  4,  6. 

187.  Problem  125.   Find  the  angle  between  the  line  [Jf  =  —  6,  3,  5 
N=  5,  -4,  1]  and  the  plane  S=0,  90°,  90°. 

188.  Problem  126.   Find  the  angle  between  the  line  \M=  0,  6,  —  1 
iV=  0,  -  5,  -  6]  and  the  plane  S  =  0°,  6,  3. 

189.  Problem  127.    To  find  the  angle  between  two  given  planes. 

A  dihedral  angle  is  measured  by  the  plane  angle  formed  by  two 
straight  lines  perpendicular  to  the  edge  at  the  same  point,  one  line 
lying  in  one  face  and  the  other  line  lying  in  the  other  face. 


Fig.  76 


68 


DESCRIPTIVE  GEOMETRY 


Analysis  1.  Pass  a  plane  perpendicular  to  the  line  of  intersec- 
tion of  the  two  given  planes.  This  plane  will  intersect  the  given 
planes  in  straight  lines  perpendicular  to  the  line  of  intersection 
at  the  same  point.  The  angle  between  these  two  lines  is  the 
required  angle. 

Analysis  2,   Assume  any  point  between  the  faces  of  the  dihedral 
angle.    Through  this  point  draw  two  straight  lines,  one  perpen- 
dicular to  one  face  and  the  other  perpendicular  to  the  other  face, 
/^y  The  angle  formed  by 

'\  ,^  these   two  lines   will  be 

the  supplement  of  the 
required  angle. 

Construction.  See  Fig. 
77.  Let  >S'  and  T  repre- 
sent the  given  planes 
which  intersect  in  the 
line  A-B, 

Following  Analysis  1, 
through  any  point  of 
a-hj^  as  c,,  draw  U-u^ 
perpendicular  to  «,-5,. 
U-u^  may  be  taken  as  the 
horizontal  trace  of  a  plane 
U  perpendicular  to  A-B 
(see  Section  45).  The  line  cut  from  >S^  by  ?7,  —  the  line  forming 
one  side  of  the  required  plane  angle,  —  pierces  H'dtd^;  the  line  cut 
from  T  hj  £/,  —  the  line  forming  the  other  side  of  the  same  angle, 
—  pierces  IT  at  g,,  and  it  only  remains  to  find  the  vertex  of  the 
angle,  which  is  the  point  in  which  U  is  intersected  by  A-B. 

U  will  cut  the  horizontal  projecting  plane  of  A-B  in  a  straight 
line  perpendicular  to  A-B^  piercing  H  at  c,,  and  intersecting  A-B 
at  the  vertex  of  the  angle  sought. 

Revolve  the  horizontal  projecting  plane  of  A-B  about  a-h^  as 
an  axis,  into  H.  The  line  a,-hjj  is  the  revolved  position  of  A-B^ 
and  c-fjifj  drawn  through  c,  perpendicular  to  a-hj^  is  the  revolved 
position  of  the  intersection  of  U  and  the  horizontal  projecting  plane 
of  A-B,    The  point /^^  is  the  revolved  position  of  the  vertex,  and 


RELATING  TO  POINT,   LINE,  AND  PLANE  69 

the  distance  c^-f^jj^  is  the  distance  of  the  vertex,  in  true  position, 
from  the  line  U~u^. 

With  the  vertex  in  true  position  revolve  TJ  about  TJ-u^  as  an 
axis,  into  H,  The  points  d^  and  e^  will  remain  stationary  and  the 
vertex  F  will  fall  at  /y,  where  c^-f^^  is  equal  to  Cj-fjjj^. 

The  angle  d^-fj^-e^  is  the  measure  of  the  required  angle. 

Cheek.    Pass  another  plane  perpendicular  to  A-B, 

190.  Problem  128.  Find  the  angle  between  the  two  planes 
S  =  -5,  140°,  40°  and  T  =  5,  30°,  120°. 

191.  Problem  129.  Find  the  angle  between  the  two  planes 
S  =  -2,  90°,  90°  and  T  =  4,  30°,  60°. 

192.  Problem  130.  Find  the  angle  between  the  two  planes 
^  =  0°,  -  6,  -  2  and  T  =  0°,  -  2,  -  5. 

193.  Problem  131.  Find  the  angle  between  the  plane  /S  =  —  4, 
30°,  60°  and  the  plane  IT. 

Suggestion.  In  this  case  the  edge  of  the  dihedral  angle  is  the 
horizontal  trace  of  the  given  plane.  The  auxiliary  plane,  which 
is  perpendicular  to  the  edge,  is  therefore  perpendicular  to  II.  Its 
horizontal  trace  will  be  perpendicular  to  the  horizontal  trace  of  the 
given  plane,  and  its  vertical  trace  will  be  perpendicular  to  G-L. 

194.  Problem  132.  Find  the  angle  between  the  plane  S  =  —  S, 
20°,  60°,  and  the  plane  V. 

195.  Problem  133.  Griven  either  trace  of  a  plane  and  the  angle 
which  the  plane  makes  with  the  corresponding  plane  of  projection ; 
required  to  determine  the  other  trace. 

Ayialysis.  Suppose  the  horizontal  trace  of  the  plane  and  the 
angle  which  the  plane  makes  with  H  are  given. 

Any  plane  perpendicular  to  this  horizontal  trace  will  cut  from 
the  plane  whose  vertical  trace  is  sought  a  straight  line,  and  from 
H  another  straight  line,  both  of  which  will  be  perpendicular  to 
the  horizontal  trace  at  the  same  point.  These  two  lines  will  form 
a  plane  angle  measuring  the  given  dihedral  angle,  and  if  revolved 
into  H  about  the  side  in  iT,  will  be  seen  in  true  size. 

Through  any  point  of  the  given  horizontal  trace  draw  two 
straight  lines  in  H^  the  first  perpendicular  to  the  trace  and  the 
second  making  an  angle  with  the  first  equal  to  the  measure  of  the 
given  dihedral  angle. 


70 


DESCRIPTIVE  GEOMETRY 


Revolve  the  plane  of  these  two  lines  about  the  first  line  as  an 
axis  until  perpendicular  to  //.  The  second  line  in  this  position 
must  be  a  line  of  the  plane  whose  vertical  trace  is  sought,  and 
must  pierce  F  at  a  point  in  this  trace. 

By  assuming  other  points  in  the  given  horizontal  trace  and 
proceeding  as  above,  other  points  in  the  required  vertical  trace 
may  be  obtained. 

Construction.  See  Fig.  78.  Let  S-s,  represent  the  given  hori- 
zontal trace,  and  let  the  angle  A  represent  the  measure  of  the 
given  angle  which  the  plane  makes  with  //. 

Through  5,,  any  point  on  S—s^^  draw  h—d,  perpendicular  to  S-s^. 
Through  hj  also  draw  b-djj^  making  the  given  angle  A  with  h^-d,. 

The    line    ^,-<i„  is,   by   analysis, 

the  revolved  position  of  a  line 

\  \        y^  of  the  plane  whose  vertical  trace 

~~wV -^  is  sought. 

Through  d^  draw  d-d'  perpen- 
dicular to  G-L ;  also  through  d, 
draw  c?,-c?^^  perpendicular  to  b-d^, 
intersecting  b^-dj^  at  dj^. 

Revolve  the  plane  of  the  three 
lines, b-df^  b^-djj^  and  d-dj^ about 
b-df  as  an  axis  until  the  plane 
occupies  its  true  position,  which 
is  perpendicular  to  H.  The  line 
d-dfj  will  take  the  position  d-d'  perpendicular  to  G-L^  and  the 
point  d„  will  take  the  position  d'  on  the  line  d-d'^  and  at  the  dis- 
tance d,-djj  below  G-L.  The  point  c?',  then,  is  the  point  in  which 
B-D^  in  true  position,  pierces  F,  and  is  therefore  a  point  in  the 
required  vertical  trace.    S-d'-s'  is  the  required  vertical  trace. 

In  case  the  given  horizontal  trace  does  not  intersect  G-L  within 
the  limits  of  the  drawing,  assume  another  point  upon  the  hori- 
zontal trace  and  proceed  as  above  to  find  another  point  in  the 
vertical  trace. 

Check.  Assume  some  point  on  the  horizontal  trace  other  than 
those  already  used,  and  proceed  as  above  to  locate  an  additional 
point  on  the  vertical  trace  already  determined. 


Fig.  78 


KELATING  TO  POINT,  LINE,  AND  PLANE  71 

196.  Problem  134.  Given  the  horizontal  trace  [S  =  —  4,  3,  0; 
s,  =  4,  6,  0]  of  a  plane  S,  and  given  the  angle  60  degrees  which  the 
plane  S  makes  with  H;  required  the  vertical  trace  of  the  plane  S. 

197.  Problem  135.  Given  the  vertical  trace  [^  =  —5,  0,  0; 
s'  =  8,  0,  5]  of  the  plane  S,  and  given  the  angle  30  degrees  which 
the  plane  S  makes  with  V;  required  the  horizontal  trace  of  the  plane  S. 

198.  Problem  136.  Given  the  horizontal  trace  [S  =  —b,  —4,  0; 
s^  =  5,  —  4,  0]  of  the  plane  S,  and  given  the  angle  60  degrees 
which  the  playie  S  makes  with  H ;  required  the  vertical  trace  of 
the  plane  S. 

199.  Problem  137.  Given  tivo  straight  lines  not  in  the  same  plane  ; 
required  to  draw  a  third  straight  line  which  shall  he  perpendicular 
to  both. 

Analysis  1.  Through  one  of  the  lines  pass  a  plane  parallel  to 
the  other  line,  and  project  the  second  line  upon  this  plane.  This 
projection  must  be  parallel  to  the  second  line  and  intersect  the 
first  line.  At  this  point  of  intersection  erect  a  straight  line  per- 
pendicular to  the  plane. 

This  line  must  be  perpendicular  both  to  the  first  line  and  to 
the  projection  of  the  second  line.  It  will  remain  in  the  project- 
ing plane  of  the  second  line  and  therefore  intersect  the  second 
line.  It  will  be  perpendicular  to  the  second  line  since  it  is  perpen- 
dicular to  the  projection  of  the  second  line  on  a  plane  to  which 
the  second  line  is  parallel.  It  is  therefore  perpendicular  both  to 
the  first  line  and  to  the  second  line. 

Analysis  2.  In  case  we  are  required  to  find  simply  the  shortest 
distance  between  two  straight  lines  not  in  the  same  plane,  we  may 
proceed  as  follows :  Through  any  point  of  one  of  the  lines  pass  a 
plane  perpendicular  to  the  other  line.  Project  the  first  line  upon 
this  plane.  The  distance  from  the  point  in  which  the  second  line 
intersects  the  plane  to  the  projection  of  the  first  line  upon  the 
plane  is  the  required  distance. 

Construction.  See  Fig.  79.  Let  M-N  and  0-P  represent  the 
given  lines. 

Following  Analysis  1,  draw  through  M-N  the  plane  S  parallel 
to  0-P.  To  do  this,  draw  through  any  point  of  M-N.,  as  2>,  the 
line  D-E  parallel  to  0-P.    Find  the  point  e,  in  which  D-E  pierces 


72 


DESCRIPTIVE  GEOMETRY 


H',  also  find  the  point  a,  in  which  M-N  pierces  IT,  thus  locating 
the  horizontal  trace  S-Sj.  Find  the  point  h'  in  which  M-N  pierces 
F;  also  find  the  point  /'  in  which  D-E  pierces  F,  thus  locating 
the  vertical  trace  S-s'. 

Project  0-P  upon  S,  To  do  this,  draw  through  any  point  of 
0-P,  as  7f,  the  line  K-KS  perpendicular  to  aS',  and  find  its  intersec- 
tion, KS^  with  S.  The  point  KS  is  one  point  in  the  projection  of 
0-P  upon  S  (see  Section  161),  and  since  0-P  is  parallel  to  S, 
KS-X  drawn  parallel  to  0-P  is  the  required  projection. 


Fig.  79 


At  the  point  X,  where  KS-X  crosses  Jf-iV,  draw  X-  Y  perpen- 
dicular to  S  (see  Section  154),  and  note  the  point  Y  in  which  it 
intersects  0-P.  X-Y  is  the  required  line,  and  if  the  true  length 
of  its  intercept  between  M-N  and  0-P  is  desired,  it  may  be  found 
by  Problem  72. 

Check.  The  points  x^  and  x'.,  which  are  found  independently, 
should  fall  in  the  same  straight  line  perpendicular  to  G-L.  The 
same  should  be  true  of  y,  and  y'. 

A  more  severe  check  would  be  made  by  drawing  the  auxiliary 
plane,  in  the  first  part  of  the  construction,  through  0-P  parallel 
to  M-N  instead  of  through  M-N  parallel  to  0-P, 


RELATING  TO  POINT,  LINE,  AND  PLANE  73 

In  this  way  the  problem  is  solved  by  a  method  which  in  theory 
is  the  same  as  that  explained  above,  but  which  in  application  is 
entirely  independent  of  it. 

In  the  above  problem,  if  one  of  the  two  given  lines  is  perpen- 
dicular to  the  plane  of  projection,  the  required  line  will  be  parallel 
to  the  plane  of  projection,  and  its  own  projection  will  pass  through 
the  point  projection  of  the  line  which  is  perpendicular  to  the  plane 
and  will  be  perpendicular  to  the  projection  of  the  other  given  line. 

If  one  of  the  two  given  lines  is  parallel  to  the  plane  of  projection, 
the  projection  of  the  required  line  upon  this  plane  will  be  perpen- 
dicular to  the  projection  of  the  parallel  line  upon  this  plane. 

200.  Problem  138.  G-iven  the  two  straight  lines  [Jf=  — 5,  3,  5 ; 
N  =  5,  3,  5]  and  [0  =  -  4,  2,  6  ;  P  =  4,  -  4,  -  3] ;  required  the 
projections  of  the  straight  line  perpendicular  to  both. 

201.  Problem  139.  G-iven  the  two  straight  lines  [M=  —  5,  3,  5; 
]Sr=  5,  3,  5]  and  [0  =  2,  -Q,  1;  F  =  2,  1,  -  6] ;  required  the 
projections  of  the  straight  line  perpendicular  to  both, 

202.  Problem  140.  Given  the  two  straight  lines  [M=  —  5,  —6, 
-  6  ;  iV^=  3,  1,  -  2]  and  [0  =  -i,  -  6,  -  3 ;  P  =  4,  -  7,  -  6] ; 
required  the  shortest  distance  between  the  two  lines. 


CHAPTER  VII 

GENERATION  AND   CLASSIFICATION  OF  LINES 

203.  Generation  of  Lines.  We  may  regard  a  line  as  the  genera- 
tion resulting  from  the  movement  of  a  point  which  in  the  course 
of  the  generation  occupies  an  infinite  number  of  consecutive  posi- 
tions at  infinitely  small  distances  apart. 

The  portion  of  the  line  generated  by  the  point  while  moving 
from  one  position  to  its  consecutive  position  is  called  an  elementary 
line^  and  while  in  theory  it  may  be  regarded  as  having  length, 
practically  speaking  it  has  none. 

204.  Classification  of  Lines.  If  the  generating  point  moves  con- 
tinuously in  one  direction,  the  line  so  generated  is  called  a  straight 
or  a  right  line. 

If  the  generating  point  is  constantly  changing  its  direction 
of  movement,  the  line  so  generated  is  called  a  curved  line,  or  a 
curve. 

When  the  generating  point  of  ,a  curved  line  confines  its  move- 
ment to  a  plane  the  line  is  called  a  curve  of  single  curvature. 

When  the  generating  point  of  a  curved  line  does  not  confine  its 
movement  to  a  plane,  the  line  is  called  a  curve  of  double  curvature. 

205.  Curves  of  Single  and  of  Double  Curvature.  The  character  of 
a  curve,  whether  of  single  or  of  double  curvature,  will  depend 
upon  the  law  which  governs  the  motion  of  the  generating  point, 
and  there  may  be  as  many  distinct  curves  as  there  are  distinct 
laws  governing  the  motion  of  a  point. 

If  the  generating  point  moves  in  a  plane  and  retains  a  constant 
distance  from  a  fixed  point  in  the  plane,  it  will  generate  a  curve 
of  single  curvature  called  the  circle. 

If  the  generating  point  moves  in  a  plane  and  in  such  a  way  that 
the  sum  of  its  distances  from  two  fixed  points  in  the  plane  is  a 
constant  quantity,  it  will  generate  a  curve  of  single  curvature 
called  the  ellipse. 

74 


GENERATION  AND  CLASSIFICATION  OF  LINES       75 

If  the  generating  point  moves  in  a  plane,  and  in  such  a  way  that 
its  distance  from  a  fixed  point  in  the  plane  is  equal  to  its  distance 
from  a  fixed  straight  line  in  the  plane,  it  will  generate  a  curve  of 
single  curvature  called  the  parabola. 

If  the  generating  point  moves  in  a  plane,  and  in  such  a  way  that 
the  difference  of  its  distances  from  two  fixed  points  in  the  plane  is 
a  constant  quantity,  it  will  generate  a  curve  of  single  curvature 
called  the  hyperbola. 

If  the  generating  point  moves  in  such  a  way  as  to  retain  a  con- 
stant distance  from  a  fixed  straight  line,  and  to  have  a  uniform 
motion  both  around  and  in  the  direction  of  the  straight  line,  it 
will  generate  a  curve  of  double  curvature  called  the  helix. 

206.  Representation  of  Curves.  Curved  lines  like  straight  lines 
may  be  represented  by  their  projections  on  IT  and  F,  and  when  so 
represented  they  are  in  general  definitely  determined. 

These  projections  are  the  loci  of  the  projections  of  the  generat- 
ing point  in  its  consecutive  positions,  and  are  usually  curved  lines. 

If  the  curve  is  of  single  curvature  and  its  plane  is  parallel  to 
the  plane  of  projection,  its  projection  upon  this  plane  is  a  curve  of 
the  same  character  and  magnitude  as  the  original  curve. 

If  the  curve  is  of  single  curvature  and  its  plane  is  perpendicular  to 
the  plane  of  projection,  its  projection  upon  this  plane  is  a  straight  line. 

The  projection  of  a  circle  upon  a  plane  to  which  the  plane  of 
the  circle  is  oblique  is  an  ellipse.* 

The  projection  of  an  ellipse  upon  a  plane  to  which  the  plane  of 
the  ellipse  is  oblique  is  either  a  circle  or  an  ellipse.* 

The  projection  of  a  curve  of  double  curvature  is  always  a  curve 
whatever  the  relation  of  the  curve  to  the  plane  of  projection.  The 
projection  of  the  helix  upon  a  plane  perpendicular  to  its  axis  is  a 
circle. 

207.  Tangents  to  Curves.  A  straight  line  is  tangent  to  a  curve 
at  a  given  point  when  it  represents  the  rectilinear  path  in  which 
the  generating  point  is  moving  at  the  instant  it  passes  through  the 
point  of  tangency. 

*The  truth  of  these  statements  will  be  more  apparent  after  the  student  has 
obtained  some  knowledge  of  the  nature  of  the  intersections  of  cylindrical  surfaces 
by  planes.    See  Chapter  XVI. 


7G  DESCRIPTIVE  GEOMETIIY 

In  Fig.  80  let  A-B-D  represent  a  curve  generated  by  a  point 
moving  under  some  law.  Suppose  that  when  the  generating  point 
reaches  the  position  i?,  the  law  under  which  it  is  moving  ceases  to 
act  and  the  point  is  allowed  to  move  freely  in  the  direction  B-E^ 
in  which  it  is  moving  at  this  instant.  The  straight  line  B-E  is 
tangent  to  the  curve  at  the  point  B. 

Or,  a  straight  line  is  tangent  to  a  curve  at  a  given 
point  when  the  line  contains  the  given  point  and  its  con- 
secutive point.  In  Fig.  81  let  B  represent  any  point  on 
the  curve  A-B-D.  Through  B  draw  any  secant,  as  B-F^ 
cutting  the  curve  at  B  and  F.  Now  if  the  point  B  remain 
fixed  and  the  point  F  be  made  to  move  along  the  curve 
toward  B^  the  secant  will  gradually  approach  the  position 
of  a  straight  line  tangent  to  the  curve  at  the  point  B. 
Finally,  when  the  point  i^  has  taken  the  position  consecu- 
tive to  B^  or  practically  the  position  B  itself,  the  line  B-F  will 
take  the  position  B-E  and  be  tangent  to  the  curve  at  the  point  B, 
Two  curves  are  tangent  to  each  other  when  they  have  two  con- 
secutive points  in  common,  or  when  they  are  both  tangent  to  the 
same  line  at  a  common  point. 


\ 


A  straight  line  which  is  tangent  to  a  curve  of 
single  curvature  lies  in  the  plane  of  the  curve. 

If  two  lines,  straight  or  curved,  are  tangent  to 
each  other,  their  projections  will  also  be  tangent 
to  each  other,  since  the  projections  of  the  two  ^  . 
consecutive  points  common  to  the  two  lines  will 
also  be  consecutive  and  be  common  to  the  two 
projections  of  these  lines. 

208.  Problem  141.    To  draw  a  rectilinear  tmigent 
to  an  ellipse  at  a  poiyit  assumed  on  the  curve.* 

Construction.^  In  Fig.  82  let  A-B  and  D-E  represent  the  axes 
of  the  ellipse,  let  F  and  F'  represent  the  foci  of  the  ellipse,  and 
let  P  represent  the  point  assumed  on  the  curve. 

*  The  metliod  of  representing  the  ellipse,  the  parabola,  and  the  hyperbola  is 
fully  explained  in  text-books  on  elementary  mechanical  drawing. 

t  The  proof  of  the  methods  here  given  for  drawing  rectilinear  tangents  to  the 
ellipse,  the  parabola,  and  the  hyperbola  may  be  found  in  text-books  on  analytic 
geometry. 


GENERATION  AND  CLASSIFICATION  OF  LINES       77 


Fk;.  82 


Draw  the  focal  radii  F-P  and  F'-P^  and  bisect  their  external 
angle  by  the  line  P-0.    The  line  P-0  is  the  required  tangent. 

209.  Problem  142.  Todraw  arec- 
tilinear  tangemt  to  a  parabola  at  a 
point  assimied  on  the  curve. 

Construction.  In  Fig.  83  let  A-C 
represent  the  axis  of  the  parabola, 
let  A-B  represent  the  directrix, 
let  D  represent  the  vertex,  and 
let  P  represent  the  point  assumed  on  the  curve. 

Draw  P-L  perpendicular  to  A-C.    Lay  off  B-0  equal  to  D-L. 

The  straight  line  0-P  through  0 
and  P  is  the  required  tangent. 

210.  Problem  143.  Todrawaree- 
tilinear  tangent  to  a  hyperhola  at  a 
point  assumed  on  the  curve. 

Construction.  In  Fig.  84  let  A-B 
represent  the  axis  of  the  hyper- 
bola, let  F  and  F'  represent  the 
foci,  let  A  and  B  represent  the 
vertices  of  the  two  branches  of 
the  curve,  and  let  P  represent  the  point  assumed  on  the  left-hand 
branch  of  the  curve. 

Draw  the  focal  radii  P-i^^  and 
P-F'  and  bisect  the  angle  F-P- 
F'  by  the  line  P-0.  The  line 
P-0  is  the  required  tangent. 

211.  Normals  to  Curves.  A 
straight  line  is  normal  to  a 
curve  at  a  given  point  when  it 
is  perpendicular  to  the  recti- 
linear tangent  to  the  curve  at 
that  point. 

A  normal  to  a  curve  of  single     ^  ^ig-  84 

curvature,  like  the  rectilinear  tangent,  lies  in  the  plane  of  the  curve. 

212.  Rectification  of  Curves.    A   curve  is  said   to  be  rectified 
when  its  length  in  linear  units  is  determined. 


Fig.  83 


78  DESCRIPTIVE  GEOMETRY 

If  a  curve  is  rolled  out  upon  a  rectilinear  tangent  to  the  curve, 
so  that  the  consecutive  points  of  the  curve  fall  consecutively  upon 
the  tangent,  that  portion  of  the  tangent  covered  between  the  first 
and  last  points  of  contact  will  represent  the  rectification  of  that 
portion  of  the  curve  between  these  same  points. 

The  work  of  rectification  is  accomplished  graphically  by  divid- 
ing the  curve  into  a  large  number  of  small  arcs,  so  small  that  the 
chords  of  the  arcs  may  for  practical  purposes  be  taken  as  equal  in 
length  to  the  arcs  themselves,  and  by  taking 
the  summation  of  these  chords. 

In  Fig.   85   let   it  be  required  to  rectify 
the  portion  A-B  of  the   curve  A-B-D.    Set 
the  dividers  at  some  sufficiently  small  distance 
and  apply  this  distance  as  a  chord  successively 
to  the  curve,  starting  with  the  point  A.    In 
this  particular  case  the  chord  is  applied  six 
_^.____________      times,  leaving  the  little  arc  Q-B  whose  chord 

'  '  '  '      '      '         'is  smaller  than  the  chord  assumed. 

Fig.  85  ^ow  apply  the  original  chord  distance  six 

times  to  the  straight  line  A-D^,  starting  with  J,,  adding  to  the  sum 
the  little  distance  6 -^^  which  is  the  chord  measure  of  the  little 
arc  Q-B  on  the  original  curve.  The  distance  A^-B,  is  the  rectified 
length  of  the  arc  A-B. 

213.  The  Helix.  The  helix  is  a  curve  of  double  curvature  gen- 
erated by  a  point  moving  uniformly  both  around  and  in  the  direction 
of  a  given  straight  line  from  which  it  retains  a  constant  distance. 

The  straight  line  around  and  along  which  the  generating  point 
moves  is  called  the  axis  of  the  helix. 

The  constant  distance  which  the  generating  point  retains  from 
the  axis  is  called  the  radius  of  the  helix. 

The  distance  over  which  the  generating  point  moves  in  the 
direction  of  the  axis  during  one  circuit  of  the  axis  is  called  the 
pitch  of  the  helix. 

If,  to  an  observer  looking  along  the  axis  in  that  direction  in 
which  the  generating  point  is  moving,  the  circuit  of  the  gener- 
ating point  about  the  axis  is  clockwise,  the  helix  is  called  a  right- 
handed  helix. 


(lENEKATKJN  A^ND   CLASSIFICATION   OF  LINES       79 


If,  under  the  same  conditions,  the  circuit  is  contra-clockwise, 
the  helix  is  called  a  left-handed  helix. 

A  helix  is  completely  known  when  the  radius,  the  pitch,  and 
the  direction  of  the  circuit  (whether  right-handed  or  left-handed) 
are  given. 

214.  To  represent  the  Helix.  See  Fig.  86.  Let  A-B  represent 
the  axis  of  the  helix,  assumed  for  convenience  perpendicular  to  H, 

With  a^  as  a  center  and  with  a-dj  as  a  radius,  equal  in  length 
to  the  radius  of  the  helix,  draw  the  circle  d^-dj-d,,-  •  — c?^^.  The 
circumference  of  this  circle  will  be  the 
horizontal  projection  of  the  helix, 
since  the  generating  point  retains  a 
constant  radial  distance  from  the  axis. 

Let  D  (df,  d')  represent  the  posi- 
tion of  the  generating  point  when 
in  H,  Beginning  with  d,  divide  the 
circumference  of  the  circle  into  any 
number,  say  eight,  equal  parts.  The 
points  of  division  are  c?,,  cZ,,,  (?,,p  •••, 
dj^.  Lay  off  upon  a'-h',  downward 
from  G-L^  a  distance  a'-e'  equal  to 
the  pitch  of  the  helix.  Divide  this 
distance  a'-e'  into  the  same  number 
of  equal  parts  as  the  circumference 
of  the  circle  has  been  divided  into,  and  through  the  points  of  divi- 
sion, 1,  2,  3,  4,  •  •  •,  e\  draw  horizontal  straight  lines. 

While  the  generating  point  is  making  one  circuit  of  the  axis 
it  moves  through  a  vertical  distance  equal  to  a'—e'.  Therefore, 
since  the  two  motions  are  uniform,  while  the  generating  point  is 
making  any  fractional  part  of  the  circuit  it  must  move  through 
the  same  fractional  part  of  the  total  vertical  distance. 

When  the  generating  point  is  on  ^  its  horizontal  projection  is 
at  d,  and  its  vertical  projection  is  at  d'. 

When  the  generating  point  has  made  one  eighth  of  its  circuit 
its  horizontal  projection  will  be  at  d,,.  Its  vertical  projection  must 
be  on  a  straight  line  through  d,,  perpendicular  to  G-L^  and  on  a 
straight  line  through  1  parallel  to  G-L^  and  therefore  at  d". 


80 


DESCRIPTIVE  GEOMETEY 


If  we  draw  vertical  lines  through  c?,,^  c?,,,,,  •  •  •,  c?^^'  ^^^^  ^ote  their 
intersections  with  the  corresponding  horizontal  lines,  we  shall 
obtain  the  points  d'",  d"'\  •••,  t?^-^,  other  points  in  the  vertical  pro- 
jection of  the  helix. 

215.  To  assume  a  Point  upon  the  Helix.  See  Fig.  86.  Assume 
the  horizontal  projection  of  the  point  anywhere  upon  the  hori- 
zontal projection  of  the  helix,  and  through  this  point  draw  a 
straight  line  perpendicular  to  G-L  to  intersect  the  vertical  pro- 
jection of  the  helix  in  the  required  vertical  projection  of  tlie  point. 

216.  To  draw  a  Rectilinear  Tangent  to  a  Helix  at  a  Point  assumed 
on  the  Curve. 

Analysis.  Since  the  generating  point  of  a  helix  retains  a  constant 
distance  from  the  axis,  the  curve  may  be  traced  upon  the  surface 
of  a  right  circular  cylinder  whose  axis  is  the  axis  of  the  helix. 


Fig.  87 


Fig.  87  represents  pictorially  a  right  circular  cylinder  with  a 
portion  of  a  helix,  a-B-C-B-E-F.,  traced  upon  the  surface. 

Assuming  the  axis  of  the  cylinder  perpendicular  to  //,  the  portion 
of  the  base  a^-b^-c^-d^-e^-f^j  represents  the  horizontal  projection 
of  the  portion  of  the  helix  in  question. 

Let  a,  represent  the  point  in  which  the  helix  pierces  H.  Let  B 
represent  the  point  consecutive  to  a,  (greatly  magnified  in  its  dis- 
tance from  a,),  and  let  h^  represent  the  horizontal  projection  of  B. 
Let  C  represent  the  point  consecutive  to  B  and  let  c^^  represent  the 
horizontal  projection  of  C 


GENERATION  AND  CLASSIFICATION  OF  LINES       81 

In  the  same  way  let  D,  E^  F^  etc.,  represent  consecutive  points 
of  the  curve  and  let  djj,  e,,,  /,,,  etc.,  represent  respectively  their 
horizontal  projections. 

Connect  a^  and  ^  by  a  straight  line ;  also  connect  a^  and  h^  by  a 
straight  line.  These  two  lines  together  with  the  horizontal  project- 
ing line  of  B  form  a  right-angled  triangle  in  which  a^-hj  is  the 
horizontal  projection  of  a-B. 

Practically  speaking,  the  two  lines  a-B  and  aj-h,  are  identical 
with  the  arcs  of  which  they  are  chords.  For  this  reason  the  angle 
B-a,-hj  measures  the  slope  of  the  helix,  that  is,  the  constant  inclina- 
tion of  the  curve  to  H^  or  the  angle  which  a  rectilinear  tangent  to 
the  helix  at  any  point  makes  with  H, 

Connect  B  and  C  by  a  straight  line  ;  also  connect  h,  and  c^^  by  a 
straight  line.  These  two  lines  together  with  the  two  projecting 
lines  B-hf  and  C-Cj^  form  a  quadrilateral  in  which  B-C  will  make 
the  same  angle  with  H  as  that  made  by  a^-B^  since  the  uniform 
motions  of  the  generatinp-  point  give  to  each  elementary  portion  of 
the  helix  the  same  inclination  to  H. 

If  the  plane  of  the  triangle  a-B-hj  be  revolved  about  B-hj  as  an 
axis  until  it  coincides  with  the  plane  of  the  quadrilateral  B-c^,^  the 
line  a-h^  will  take  the  position  a^^-h^^  which  is  a  continuation  of 
Cj-hj\  and  the  line  a-B  will  take  the  position  a^-B^  which  is  a 
continuation  of  C-B, 

If  the  plane  of  the  triangle  a^-C-c^^  be  revolved  about  C-c^^  as 
an  axis  until  it  coincides  with  the  plane  of  the  quadrilateral  C-c?,,, 
the  line  a^-h^-c^,  will  take  the  position  a^^-h^-c^^^  which  is  a  con- 
tinuation of  d^-G^^ ;  and  the  line  a^-B-C  will  take  the  position 
a„-C,  which  is  a  continuation  of  D-C. 

The  line  a^^-B  is  tangent  to  the  helix  at  the  point  B,  since  it 
contains  B  and  its  consecutive  point  C.  The  line  (i,-hf  is  the  hori- 
zontal projection  of  this  tangent  and  is  itself  tangent  to  the  hori- 
zontal projection  of  the  helix  at  the  points,.  The  tangent  a^^-B 
pierces  H  a.t  a,^^  at  a  distance  from  5,- equal  to  the  rectification  of 
the  arc  a-h^. 

Again,  a^^-C  is  tangent  to  the  helix  at  the  point  C,  since  it  con- 
tains C  and  its  consecutive  point  D.  The  line  a^j-hj-Cj^  is  the 
horizontal  projection  of  this  tangent  and  is  itself  tangent  to  the 


82 


DESCRIPTIVE   GEOMETRY 


horizontal  projection  of  the  helix  at  the  point  c^^.  The  tangent 
ci,,—C  pierces  i:r  at  «,,,,  at  a  distance'  from  c^j  equal  to  the  rectification 
of  the  arc  a-h-Cf^. 

That  portion  of  the  tangent  a,-B  between  the  points  a,,  and  B 
is  the  rectified  length  of  the  curve  a-B,  and  that  portion  of  the 
tangent  a^^-C  between  the  points  a,^,  and  C  is  the  rectified  length 
of  the  curve  a^-B-C. 

We  have  now  considered  three  consecutive  points,  a„  B,  and  C\ 
of  the  helix,  and  it  will  be  evident  that  the  same  process  of  reason- 
ing may  be  applied  indefinitely  to  the  consecutive  points  of  the 
curve. 

We  may  conclude,  then,  that  the  horizontal  projection  of  a  recti- 
linear tangent  to  a  helix  at  any  point  on  the  curve,  provided  the 
axis  of  the  helix  is  perpendicular  to  If^  will  be  tangent  to  the  hori- 
zontal projection  of  the  helix  at  the  horizontal  projection  of  the- 
point  of  tangency ;  also  that  the  tangent  itself  will  pierce  H  upon 
the  horizontal  projection  of  the  tangent  and  at  a  distance  from  the 
horizontal  projection  of  the  point  of  tangency  equal  to  the  rectifi- 
cation of  that  portion  of  the  horizontal 
projection  of  the  helix  between  the  point 
in  which  the  helix  pierces  H  and  the 
horizontal  projection  of  the  point  of 
tangency. 

By  use  of  these  principles  we  can  draw 
the  two  projections  of  a  rectilinear  tan- 
gent to  the  helix  when  the  axis  of  the 
helix  is  assumed  perpendicular  to  //. 

Construction.  Let  the  helix  be  repre- 
sented as  in  Fig.  88.  Assume  any  point, 
as  E,  upon  the  curve  (see  Section  215). 
At  e,  draw  e^-f^  tangent  to  the  horizontal 
projection  of  the  helix.  Upon  this  tan- 
gent lay  off  from  e,  a  distance  e^-ff  equal  to  the  rectification  of 
the  arc  e-d^,  where  d^  is  the  point  in  which  the  helix  pierces  H, 
and  where  e,  is  the  horizontal  projection  of  the  point  of  tangency. 
The  line  e-f,  is  the  horizontal  projection  of  the  required  tangent, 
and  /,  is  the  point  in  which  this  tangent  pierces  H.    The  vertical 


GENEKATION  AND  CLASSIFICATION  OF  LINES       8 


o 


projection  of  F  is  at/',  on  G-L,  and /'-e'  is  the  vertical  projection 
of  the  required  tangent. 

By  this  process  the  position  of  the  vertical  projection  of  the  tan- 
gent to  the  helix  is  determined  by  two  points,  e'  and  /',  and  the 
unsatisfactory  task  of  drawing  a  rectilinear  tangent  to  an  irregular 
curve  is  avoided. 

217.  Problem  144.  Represent  a  left-handed  helix  whose  radius  is 
4  and  whose  pitch  is  8. 

218.  Problem  145.  Find  the  point  in  which  the  right-handed  helix 
whose  radius  is  4  and  pitch  10  intersects  a  horizontal  plane  4  above  H, 

219.  Problem  146.  Fiyid  the  point  in  which  the  right-handed 
helix  whose  radius  is  4  and  pitch  9  intersects  a  plane  containing  the 
axis  and  making  an  angle  of  45  degrees  with  V, 


CHAPTER  VIII 

GENERATION  AND   CLASSIFICATION   OF  SURFACES 

220.  Generation  of  Surfaces.  We  may  regard  a  surface  as  the 
generation  resulting  from  the  movement  of  a  line  which  in  the 
course  of  the  generation  occupies  an  infinite  number  of  consecu- 
tive positions  at  infinitely  small  distances  apart.  The  moving  line 
is  called  the  generatrix^  and  the  various  positions  which  it  occu- 
pies during  the  generation  are  called  elements  of  the  surface. 

The  portion  of  the  surface  generated  by  the  line  while  moving 
from  one  position  to  its  consecutive  position  is  called  an  elementary 
surface^  and  while  in  theory  the  distance  between  the  two  con- 
secutive elements  must  be  taken  into  account,  practically  speak- 
ing the  two  elements  may  be  regarded  as  one  and  the  same. 

221.  Classification  of  Surfaces.  The  character  of  the  surface 
generated  will  depend  both  upon  the  character  of  the  generatrix 
and  upon  the  nature  of  its  motion. 

Depending  upon  the  character  of  the  generatrix  we  have  two 
classes  of  surfaces:  first,  those  which  are  generated  by  straight 
lines,  or  those  which  have  rectilinear  elements ;  and  second,  those 
which  are  generated  by  curved  lines,  known  as  surfaces  of  double 
curvature. 

Depending  upon  the  nature  of  the  motion  of  a  rectilinear  gen- 
eratrix w^e  have  (1)  the  plane  which  may  be  generated  by  one 
straight  line  moving  in  such  a  way  as  to  touch  another  straight 
line  and  always  remaining  parallel  to  its  first  position ;  (2)  the 
single  curved  surface  which  may  be  generated  by  a  straight  line 
moving  in  such  a  way  that  any  two  of  its  consecutive  positions 
shall  be  in  the  same  plane ;  and  (3)  the  warped  surface  which  may 
be  generated  by  a  straight  line  moving  in  such  a  way  that  no  two 
of  its  consecutive  positions  shall  be  in  the  same  plane. 

222.  Single  Curved  Surfaces.  When  the  rectilinear  generatrix 
moves  in  such  a  way  that  all  its  positions  are  parallel  but  no  three 

84 


CLASSIFICATION  OF   SURFACES  85 

of  its  consecutive  positions  lie  in  the  same  plane,  the  surface  is  a 
cylhidrical  surface. 

When  the  rectilinear  generatrix  moves  in  such  a  way  that  all  its 
positions  pass  through  a  common  point  and  no  three  of  its  con- 
secutive positions  lie  in  the  same  plane,  the  surface  is  a  conical 
surface. 

When  the  rectilinear  generatrix  moves  in  such  a  way  that  con- 
secutive positions  intersect  two  and  two  and  at  the  same  time  in 
such  a  way  that  no  three  consecutive  positions  lie  in  the  same 
plane,  the  surface  is  a  convolute. 

223.  Surfaces  of  Revolution.  Surfaces  of  revolution  are  those 
which  may  be  generated  by  the  revolution  of  a  line  about  a 
straight  line  as  an  axis. 

If  the  generatrix  is  a  straight  line  and  is  parallel  to  the  axis, 
the  surface  is  a  cylindrical  surface  of  revolution^  —  a  single  curved 
surface. 

If  the  generatrix  is  a  straight  line  and  intersects  the  axis 
obliquely,  the  surface  is  a  conical  surface  of  revolution.,  —  a  single 
curved  surface. 

If  the  generatrix  is  a  straight  line  and  does  not  lie  in  the  same 
plane  with  the  axis,  the  surface  is  a  warped  surface  of  revolution^ 
since  from  the  nature  of  the  generation  consecutive  elements  of 
the  surface  cannot  lie  in  the  same  plane. 

If  the  generatrix  is  a  curved  line,  as  it  will  be  in  all  cases  save 
those  mentioned  above,  the  surface  will  be  one  of  double  curva- 
ture, or  a  double  curved  surface  of  revolution. 

224.  Double  Curved  Surfaces  of  Revolution.  If  the  generatrix  of 
a  double  curved  surface  of  revolution  is  the  circumference  of  a 
circle  and  the  axis  is  a  diameter  of  the  circle,  the  surface  generated 
is  that  of  a  sphere. 

If  the  generatrix  is  the  curve  of  an  ellipse  and  the  axis  is  one  of 
the  axes  of  the  ellipse,  the  surface  generated  is  that  of  the  ellipsoid 
of  revolution.  The  ellipsoid  of  revolution  is  called  a  prolate  or  an 
ohlate  spheroid  according  as  the  long  or  the  short  axis  is  used. 

If  the  generatrix  is  the  curve  of  a  parabola  and  the  axis  is  the 
axis  of  the  parabola,  the  surface  generated  is  that  of  the  paraboloid 
of  revolution. 


86  DESCiilPTIVE  GEOMETRY 

If  the  generatrix  is  the  curve  of  a  hyperbola  and  the  axis  is  one 
of  the  axes  of  the  hyperbola,  the  surface  generated  is  that  of  the 
hyperholoid  of  revolution. 

225.  Representation  of  Surfaces.  Surfaces  which  exist  within 
definite  limitations  are  usually  represented  by  projecting  upon 
H  and  V  the  limiting  lines  as  seen  from  the  two  principal  stand- 
points of  projection.  Other  surfaces  are  too  irregular  in  their 
formation  to  be  represented  in  this  way,  and  all  that  is  attempted 
is  to  represent  by  projection  a  sufficient  number  of  the  elements  of 
the  surface  to  reveal  tlie  character  of  some  small  portion  of  the 
surface  under  consideration. 

226.  Tangents  to  Surfaces.  A  straight  line  is  tangent  to  a  sur- 
face at  a  given  point  wlien  it  is  tangent  to  a  line  of  the  surface  at 
that  point. 

A  plane  is  tangent  to  a  surface  at  a  given  point  when  it  con- 
tains all  the  rectilinear  tangents  to  the  surface  at  that  point.  In 
other  words,  if  a  plane  is  tangent  to  a  surface,  and  any  cutting 
plane  be  passed  through  the  point  of  tangency,  the  cutting  plane 
will  cut  from  the  surface  a  line,  and  from  the  tangent  plane  a 
straight  line,  tangent  to  the  first  line  at  the  point  of  tangency. 

Therefore,  to  draw  a  plane  tangent  to  a  surface  at  a  given  point, 
draw  two  rectilinear  tangents  to  the  surface  at  this  point  and 
determine  their  plane. 

If  the  surface  has  rectilinear  elements,  the  tangent  plane  must 
contain  the  rectilinear  element  passing  through  the  point  of  tan- 
gency, since  the  rectilinear  tangent  Xo  a  rectilinear  element  is  the 
element  itself.     This  element  is  the  element  of  tangency. 

If  the  surface  is  of  single  curvature,  we  may  say  that  a  plane  is 
tangent  to  the  surface  at  a  given  point  when  it  represents  the 
plane  in  which  the  generatrix  is  moving  at  the  instant  in  which  it 
passes  through  the  point  of  tangency. 

For  this  reason  we  may  say  that  a  plane  is  tangent  to  a  single 
curved  surface  at  a  given  point  when  it  contains  both  the  element 
through  the  point  of  tangency  and  its  consecutive  one.  If  through 
the  element  containing  the  point  of  tangency  we  pass  a  secant 
plane,  it  will  cut  the  surface  in  two  distinct  lines,  one  of  which  is 
the  element  through  the  point  of  tangency  and  the  other  another 


CLASSIFICATION  OF   SUEFACES  87 

line  somewhat  removed  from  the  first.  If  this  secant  plane  be 
revolved  about  the  element  of  tangency  as  an  axis  toward  the 
position  of  the  tangent  plane,  the  element  of  tangency  will  remain 
stationary  and  the  second  line  will  gradually  approach  the  position 
of  the  first ;  and  when  the  second  line  takes  the  position  consecu- 
tive to  the  element  of  tangency,  or  practically  the  position  of  the 
element  of  tangency  itself,  the  secant  plane  will  take  the  position 
of  the  tangent  plane. 

If  a  plane  is  tangent  to  a  single  curved  surface  at  a  given  point, 
it  will  be  tangent  to  the  surface  all  along  the  rectilinear  element 
containing  the  point  of  tangency.  For  if  through  any  point  of 
this  element  a  cutting  plane  be  passed  oblique  to  the  elements,  it 
will  cut  from  the  consecutive  element  (which  is  also  a  line  of  the 
tangent  plane)  a  point  consecutive  to  the  assumed  point. 

These  two  consecutive  points  lie  in  the  tangent  plane  and  at  the 
same  time  lie  on  the  line  cut  from  the  surface  by  the  oblique  plane. 
A  straight  line  through  these  two  points  is  tangent  to  the  line  cut 
from  the  surface  at  the  assumed  point,  and  lies  in  the  tangent 
plane.  The  tangent  plane  is  then  tangent  to  the  surface  at  this 
point,  since  it  contains  two  rectilinear  tangents  to  the  surface  at 
this  point. 

If  then  a  plane  is  tangent  to  a  single .  curved  surface,  any  plane 
passed  oblique  to  the  element  of  tangency  will  cut  from  the  tan- 
gent plane  a  straight  line  wliich  will  be  tangent  to  the  line  cut 
from  the  surface  at  the  point  where  the  element  of  tangency  inter- 
sects the  oblique  plane. 

Two  surfaces  are  tangent  to  each  other  when  they  are  both 
tangent  to  the  same  surface  at  a  common  point,  or  when  planes 
passed  through  their  point  of  contact  cut  from  the  two  surfaces 
lines  which  are  tangent  to  each  other  at  the  point  of  contact. 

227.  Normals  to  Surfaces.  A  straight  line  is  normal  to  a  surface 
at  a  given  point  when  it  is  perpendicular  to  the  plane  which  is 
tangent  to  the  surface  at  that  point. 

A  plane  is  normal  to  a  surface  at  a  given  point  when  it  contains 
the  rectilinear  normal  to  the  surface  at  that  point. 

Evidently  there  can  be  but  one  rectilinear  normal  to  a  surface 
at  a  given  point,  but  an  infinite  number  of  plane  normals. 


88  DESCRIPTIVE  GEOMETRY 

228.  The  Development  of  Surfaces.  By  the  development  of  a 
surface  we  mean  its  rectification,  or  its  measure  or  appearance 
when  laid  out  on  a  plane. 

Just  as  a  curved  line  is  rectified  by  rolling  the  curve  out  on 
a  rectilinear  tangent  to  the  curve,  so  a  surface  is  developed  by 
rolling  the  surface  out  on  a  plane  tangent  to  the  surface. 

In  cases  of  prisms  and  pyramids,  or  of  any  surfaces  made  up 
of  plane  faces,  the  plane  of  one  of  the  faces  is  taken  as  the  plane 
of  development,  and  the  successive  faces  are  brought  into  coin- 
cidence with  the  plane  of  development  by  revolving  them  one 
after  another  about  the  edges  as  axes. 

In  cases  of  cylinders  and  cones,  or  other  surfaces  of  single 
curvature,  a  plane  tangent  to  the  surface  along  some  element  is 
taken  as  the  plane  of  development,  and  the  surface  is  tlien  rolled 
out  element  after  element,  peeling  off,  as  it  were,  the  outer 
coating  of  the  surface  and  spreading  it  out  on  the  plane. 

The  development  of  a  surface  may  be  used  as  a  templet,  or 
pattern,  for  cutting  out  a  plane  surface  form,  which  by  a  process 
converse  to  that  employed  in  development  may  be  made  to  take 
the  shape  of  the  original  surface. 


CHAPTER  IX 

REPRESENTATION   OF   SURFACES  WITH  PLANE  FACES 

229.  Surfaces   consisting    of    Plane    Faces.    Such   surfaces   are 
usually  placed,  for  ease  of  representation,  in  such  a  position  that 


b, 

9f 

k 

d, 

fr 

h 

a, 

^t 

a' 

b' 

d' 

e' 

f 

y 

V 

/ 

/n 

> 

t 

a,|^ 

<  / 

ri,^ 

H 

u 

a' 

16' 
1 

e' 

d' 

r 

1 
|9' 

V 

k' 

Fig.  89 


Fig.  90 


their  principal  faces  are  either  perpendicular  to  or  parallel  to  the 
planes  of  projection. 

Fig.  89  represents  a  cube  in  the  third  quadrant  with  two  of 
its  faces  parallel  to  H  and  with  two  of  its  faces  parallel  to  F". 

h 


h 


9f 


d, 


/-Y 

Or 

b' 

d' 

f 

9' 

k> 

V 

G 


(U 

K 

> 

d, 

M^^ 

i         i 

a' 

b' 

e' 

d' 

L 

9' 

I' 

k' 

Fig.  91 


Fig.  92 


Fig.  90  represents  a  cube,  two  of  whose  faces  are  parallel  to  H 
but  whose  vertical  faces  are  oblique  to  V, 


89 


90 


DESCRIPTIVE   GEOMETRY 


Fig.  91  represents  a  rectangular  prism  with  bases  parallel  to  H 
and  with  two  faces  parallel  to  V. 

Fig.  92  represents  a  rectangular  prism  whose  edges  are  perpen- 
dicular to  H  and  whose  faces  are  oblique  to  V. 


Fig.  93 


Fig.  94 


Fig.  93  represents  a  rectangular  prism  whose  edges  are  perpen- 
dicular to  V  and  whose  faces  are  oblique  to  H, 


Fig.  95 


Fig.  96 


Fig.  94  represents  a  hexagonal  prism  with  bases  parallel  to  H 
and  with  edges  perpendicular  to  H. 

Fig.  95  represents  an  oblique  hexagonal  prism  with  bases  parallel 
to  //. 


SURFACES  WITH  PLANE  FACES 


91 


Fig.  96  represents  a  square  pyramid  with  base  parallel  to  H  and 
with  axis  perpendicular  to  H. 

Fig.  97  represents  a  pentagonal  pyramid  with  base  parallel  to 
H  and  with  axis  perpendicular  to  H. 

Fig.  98  represents  the  frustum  of  a  square  pyramid  with  bases 
parallel  to  -^. 


Fi(i.  97 


Fig.  98 


230.  Shade  Lines.  Objects  in  nature,  as  a  rule,  are  exposed  to 
some  source  of  light  so  that  some  portions  of  their  surface  are  in 
the  light  and  other  portions  are  in  the  dark. 

The  sun  is  usually  taken  as  the  source  of  light,  and  on  account 
of  its  great  distance  from  the  earth  we  may  safely  consider  such 
solar  rays  as  fall  upon  terrestrial  objects  of  finite  dimensions  as 
parallel. 

The  source  of  light  is  conventionally  assumed  in  such  a  posi- 
tion that  the  rays  shall  be  parallel  to  that  diagonal  of  a  cube  (the 
cube  resting  on  H  in  the  first  quadrant  with  one  face  coincident 
with  V)  which  slopes  downward  to  the  right  toward  V. 

Under  these  conditions  the  horizontal  and  vertical  projections 
of  a  ray  of  light  will  each  make  an  angle  of  45  degrees  with  G-L. 

When  the  position  of  the  object  is  definitely  known  it  will  be 
easy  to  determine  which  portions  of  the  surface  are  in  the  light 
and  which  portions'  are  in  the  dark. 


92  DESCRIPTIVE   GEOMETRY 

Those  lines  on  a  surface  which  separate  light  portions  from 
dark  portions  are  called  shade  lines,  and  in  the  drawing,  for  pur- 
poses of  clearness,  are  represented  a  little  heavier  than  the  ordi- 
nary line. 

In  Fig.  89,  remembering  the  direction  taken  by  rays  of  light, 
it  is  evident  that  the  upper,  the  left-hand,  and  the  front  faces  of 
the  cube  are  in  the  light  and  that  all  the  others  are  in  the  dark. 

Therefore  the  shade  lines  in  this  case  are  B-D,  D-E,  E-L,  L-F, 
F-G,  and  G-B,  and  are  therefore  represented  by  heavy  lines. 

In  Fig.  90  the  shade  line  B-G  is  invisible  and  is  therefore 
represented  by  a  heavy  broken  line. 

When  two  lines  of  the  surface,  one  a  shade  line  and  the  other 
an  ordinary  line,  are  projected  upon  the  same  line,  it  is  customary 
to  give  preference  to  that  line  which  is  visible.  For  example,  in 
Fig.  89  the  lines  A-E  and  F-L  have  a  common  horizontal  pro- 
jection ttf-e^,  but  the  line  A-E  is  the  visible  line,  and  inasmuch 
as  it  is  not  a  shade  line  the  projection  a-e,  is  made  light. 

In  Fig.  90,  B-D  and  G-K  have  a  common  horizontal  projection, 
but  since  B-D  is  the  visible  line  and  is  a  shade  line,  b-d,  is  made 
a  heavy  line. 

Following  the  directions  now  given,  it  will  be  easy  to  determine 
the  shade  lines  in  the  remaining  diagrams. 


CHAPTER  X 

REPRESENTATION  OF  SINGLE  CURVED  SURFACES 

231.  Cylindrical  Surfaces  and  the  Cylinder.  The  cylindrical  sur- 
face is  a  single  curved  surface  which  may  be  generated  by  a 
moving  straight  line  which  during  the  movement  touches  a  given 
curved  line  and  always  remains  parallel  to  its  first  position  (see 
Section  222). 

The  generating  line  is  called  the  generatrix^  the  curved  line  is 
called  the  directrix^  and  the  various  positions  occupied  by  the 
generatrix  are  called  the  rectilinear  elements  of  the  surface. 

The  intersection  of  a  cylindrical  surface  by  any  plane  which  is 
not  parallel  to  the  elements  is  called  a  section^  or  base,  of  the  sur- 
face. When  this  plane  is  taken  perpendicular  to  the  elements,  the 
section  is  called  a  right  section^  and  according  to  the  nature  of 
this  section  cylindrical  surfaces  are  classified  as  circular^  elliptical, 
parabolic,  hyperbolic,  etc. 

The  section  of  a  cylindrical  surface  made  by  the  plane  H  will 
often  serve  as  a  convenient  base. 

A  plane  parallel  to  the  elements  of  a  cylindrical  surface  and 
cutting  the  surface  will  cut  the  surface  in  elements,  since  all  the 
elements  of  such  surfaces  are  parallel. 

If  the  base  of  a  cylindrical  surface  is  a  closed  curve,  like  a  circle 
or  an  ellipse,  the  space  inclosed  by  the  cylindrical  surface  is  called 
a  cylinder. 

The  straight  line  through  the  center  of  the  base  and  parallel  to 
the  elements  of  the  surface  is  called  the  axis  of  the  cylinder. 

If  one  of  two  parallel  straight  lines  is  revolved  about  the  other 
as  an  axis,  the  surface  so  generated  is  cylindrical,  and  the  cylinder 
so  generated  is  a  cylinder  of  revolution  whose  right  section  is  a 
circle. 

232.  To  represent  the  Cylinder.  The  cylinder  is  usually  rep- 
resented by  projecting  one  of  its  bases,  which  may  be  assumed 

03 


94 


DESCRIPTIVE   GEOMETRY 


anywhere  but  which  must  not  be  regarded  as  limiting  the  surface, 
and  by  projecting  its  extreme  elements,  that  is,  those  elements  which 
from  the  observer's  position  appear  to  limit  the  surface. 

Case  1.    To  represent  a  circular  cylinder  whose  axis  is  perpen- 
dicular to  H  and  whose  bases  are  right  sections.    See  Fig.  99. 

Case  2.    To  represent  a  circular  cylinder  whose  axis  is  perpen- 
dicular to  V  and  whose  bases  are  right  sections.    See  Fig.  100. 

Case  3.  To  represent  the  cylinder  tvhen  its 
base  is  on  H.  See  Fig.  101.  Let  the  circle 
whose  center  is  a^  represent  the  base  on  H^ 
and  let  A-B  represent  the  axis.  Tangent  to 
the  circular  base  and  parallel  to  a-b^  draw 
the  straight  lines  d,-e,  and  f^-g,.  These  lines 
represent  the  horizontal  projections  of  the 
extreme  or  limiting  elements,  as  seen  from 
the  observer's  standpoint  while  projecting  on 
i/,  since  no  elements  can  occupy  positions  at 
greater  distance  from  the  horizontal  project- 
ing plane  of  A-B.  The  vertical  projections  of  I)  and  F  are  d'  and/' 
respectively.  Through  d'  and  /'  and  parallel  to  a'-b'  draw  the 
straight  lines  d'-e'  and  f-g\  to  represent  the  vertical  projections  of 
these  elements. 

Tangent  to  the  circular  base  and  perpendicular  to  G-L  draw  the 
straight  lines  k-k'  and  m-m'.    Through  k'  and  m'  and  parallel  to 


Fig.  99 


m-n 


a!—V  draw  the  straight  lines  k^-V  and 
These  lines  represent  the  vertical  projections 
of  the  extreme  or  limiting  elements,  as  seen 
from  the  observer's  standpoint  while  project- 
ing on  F,  since  no  elements  can  occupy  posi- 
tions at  greater  distance  from  the  vertical 
projecting  plane  of  A-B.  Through  k^  and  m^ 
and  parallel  to  a-b^  draw  the  straight  lines 
k^-l^  and  m-n^^  to  represent  the  horizontal 
projections  of  these  elements. 

The  outline  e^-d-k^-f-g^  represents  the  horizontal  projection 
of  the  cylinder,  and  the  outline  V-k'-m'-7i'  represents  its  vertical 
projection. 


Fig.  100 


SINGLE  CUKYED   SURFACES 


95 


Fig.  101 


It  will  be  evident  that  if  the  dimensions  or  position  of  the 
cylinder  be  changed,  the  projections  of  the  cylinder  will  also  be 
changed.  It  will  also  be  evident  that  a  cylinder  of  definite  dimen- 
sion and  position  will  have 
definite  projections  fully 
determining  the  cylinder. 

It  must  be  remembered  that 
in  assuming  the  circular  base 
and  the  inclination  of  the  axis, 
both  at  random,  we  have  rep- 
resented a  cylinder  at  random 
and  not  one  of  definite  dimen- 
sion and  position. 

Case  4.    To    represent   the 
cylinder  when  the  plane  of  its 
base  is  oblique  to  H  but  perpen- 
dicular to  V.    In  Fig.  102  the  circle  whose  center  is  a,  represents 
the   horizontal  projection  of   the  base ;    the  straight   line    k'-m' 

included  between  the  two  vertical 
lines  k,-k'  and  m—m\  each  drawn 
tangent  to  the  circle  whose  center 
is  a,,  represents  the  vertical  pro- 
jection of  the  base,  and  A-B  rep- 
resents the  axis. 

The  horizontal  projections  of 
the  extreme  elements  D-E  and 
F-G  can  now  be  determined  as 
in  the  previous  case.  The  lines 
k'-V  and  m'-n'  represent  the  ver- 
tical projections  of  the  highest 
and  lowest,  and  therefore  the  ex- 
treme, elements  as  seen  from  the 
observer's  standpoint  while  pro- 
jecting on  V. 

233.  To  assume  a  Rectilinear  Element  of  the  Cylinder.  Let  the 
cylinder  be  represented  with  a  circular  base  on  H^  as  shown  in 
Fig.  103.    Assume  any  point,  as  E^  in  the  circumference  of  the 


Fig.  102 


96 


DESCRIPTIVE  GEOMETRY 


Fig.  103 


base.    Through  e^  draw  e,-o,  parallel  to  a,-^,,  and  through  e'  draw 
e'-o'  parallel  to  a'-h'.    E-0  is  the  required  element. 

234.  To  assume  a  Point  upon  the  Surface  of  the  Cylinder.  Let 
the  cylinder  be  represented  as  in  Fig.  103.    Assume  at  random  the 

horizontal  projection  o,  of  the 
point.  Draw  o-d,-ef  parallel  to 
a^-bf,  to  represent  the  horizontal 
projection  of  the  element  con- 
taining the  point  0.  This  ele- 
ment pierces  //  either  at  d,  or  at 
e^  according  as  0  is  assumed  on 
the  upper  or  on  the  lower  surface 
of  the  cylinder.  Assuming  the 
point  on  the  lower  surface  of  the 
cylinder,  the  element  in  question 
will  be  vertically  projected  at 
e'-o\  and  the  point  0  must  be 
vertically  projected  at  o'  on  the  straight  line  through  6>,  perpen- 
dicular to  G-L. 

235.  Problem  147.  G-iven  a  cylinder  whose  axis  is  oblique  to  H 
and  V  and  whose  base  is  a  circle  in  V;  required  to  draw  the  two 
projections  of  the  cylinder  and  to  assume  a  point  upon  the  surface. 

236.  Problem  148.  Given  a  cylinder  whose  right  section  is  a  circle 
and  whose  axis  is  in  such  a  position  that  its  horizontal  projection  is 
inclined  30  degrees  to  G-L  and  its  vertical  projection  is  inclined 
60  degrees  to  G-L;  required  to  draw  the  two  projections  of  the 
cylinder  and  to  assume  a  point  upon  the  surface. 

237.  Problem  149.  Given  a  cylinder  whose  right  section  is  a  circle 
and  whose  axis  is  parallel  to  G-L;  required  to  draw  the  two  projec- 
tions of  the  cylinder  and  to  assume  a  point  upon  the  surface. 

238.  Problem  150.  Given  a  cylinder  whose  base  is  a  circle  on  If  and 
whose  axis  is  in  a  profile  plane  and  oblique  to  H ;  required  to  draw  the 
two  projections  of  the  cylinder  and  to  assume  a  point  upon  the  surface. 

239.  Conical  Surfaces  and  the  Cone.  The  conical  surface  is  a 
single  curved  surface  which  may  be  generated  by  a  moving  straight 
line  which  during  the  movement  touches  a  given  curved  line  and 
constantly  passes  through  a  given  fixed  point  (see  Section  222). 


SINGLE  CURVED  SURFACES 


97 


The  generating  line  is  called  the  generatrix^  the  curved  line  is 
called  the  directrix,  the  fixed  point  is  called  the  vertex,  and  the 
various  positions  occupied  by  the  generatrix  are  called  rectilinear 
elements  of  the  surface. 

It  is  evident  from  the  nature  of  the  generation  of  the  conical 
surface  that  there  will  be  generated  simultaneously,  on  opposite 
sides  of  the  vertex,  two  equivalent  portions  of  the  surface.  These 
portions  of  the  surface  are  called  nappes  of  the*  surface. 

It  is  also  evident  that  if  the  vertex  be  removed  to  an  infinite 
distance  from  the  directrix,  the  conical  surface  will  become  cylin- 
drical. 

The  intersection  of  a  conical  surface  by  any  plane  not  con- 
taining the  vertex  is  called  a  section,  or  base,  of  the  surface.  When 
this  plane  is  taken  perpendicular  to  the  axis*  the  section  is  called 
a  right  section,  and  conical  surfaces  are  classified  according  to 
the  nature  of  this  section  as  circular,  elliptical,  parabolic,  hyper- 
bolic, etc. 

The  section  of  a  conical  surface  made  by  the  plane  H  will  often 
serve  as  a  convenient  base. 

A  plane  containing  the  Vertex  of  a  conical  surface  and  inter- 
secting the  surface  will  cut  the  surface  in  elements,  since  all  the 
elements  of  such  surfaces  pass  through  the 
vertex. 

If  the  base  of  a  conical  surface  is  a  closed       *'^ 
curve,  like  a  circle  or  an  ellipse,  the  space 
inclosed  by  the   conical  surface  is  called  a  ^. 
cone,  and  the   straight  line   connecting    the 
center  of  the  base  with  the  vertex  is  called 
the  axis  of  the  cone. 

240.  To  represent  the  Cone.  A  cone  is  usu- 
ally represented  by  projecting  the  vertex,  a 
base,  and  the  extreme  or  limiting  elements. 

Case  1.  To  represent  a  circular  cone  whose 
axis  is  perpendicular  to  H  and  whose  base  is  a  right  section.  See 
Fig.  104,  in  which  A-B  represents  the  axis  of  the  cone  and  in  which 
F-D-G-E  represents  the  circular  base. 

*  For  definition  of  axis,  see  few  lines  below. 


Fig.  104 


98 


DESCEIPTIVE   GEOMETKV 


Case  2.    To  represent  a  circular  cone  whose  axis  is  perpendicular 
to  V  and  whose  base  is  a  right  section.    See  Fig.  105. 

Case  3.  To  represent  the  cone  when  its  base  is  on  If  and  its  axis 
is  oblique  to  H  and  V.    See  Fig.  106. 

Let  the  circle  whose  center  is  a,  represent  the  base  on  i/,  let  B 
represent  the  vertex,  and  let  A-B  represent  the  axis. 

Through  the  point  6,  and  tangent  to  the 
circle  whose  center  is  a,  draw  b-d,  and  b-e^. 
These  lines  represent  the  horizontal  projec- 
tions of  the  extreme  or  limiting  elements 
as  seen  from  the  observer's  standpoint  while 
projecting  on  H^  since  no  elements  can  occupy 
positions  at  greater  distance  from  the  hori- 
zontal projecting  plane  of  A-B.  The  vertical 
projections  of  D  and  E  are  d'  and  e'  respec- 
tively, and  the  lines  b'-d'  and  b'-e'  are  the 
vertical  projections  of  these  elements, 
^i^-  1^^  Tangent  to  the  circular  base  and  perpen- 

dicular to  G-L  dr^wf-f  and  g-g'.  Connect/'  and  b\  also  connect 
g'  and  b',  by  straight  lines.  These  lines  represent  the  vertical  pro- 
jections of  the  extreme  or  limiting  elements  as  seen  from  the 
observer's  standpoint  while 
projecting  on  F,  since  no  ele- 
ments can  occupy  positions  at 
greater  distance  from  the  ver- 
tical projecting  plane  of  A-B. 
The  lines /,-/>,  and  g-b,  repre- 
sent the  horizontal  projections 
of  these  elements. 

The  outline   b-d^-g-e—b^ 
represents  the  horizontal  pro- 
jection  of   the   cone,   and  the  ^^^'-  ^^^ 
outline  V-f-g^-V  represents  the  vertical  projection  of  the  cone. 

It  will  be  evident  that  if  the  dimensions  or  position  of  the  cone 
be  changed,  the  projections  of  the  cone  will  also  be  changed.  It 
will  also  be  evident  that  a  cone  of  definite  dimension  and  posi- 
tion will  have  definite  projections  fully  determining  the  cone. 


SINGLE  CURVED   SUEFACES 


99 


It  must  be  remembered  that  in  assuming  the  circular  base  and 
the  vertex,  both  at  random,  we  have  represented  a  cone  at  random 
and  not  one  of  definite  dimension 
and  position. 

Case  4.  To  represent  the  cone 
tvJien  the  plane  of  its  base  is  oblique  to 
H  but  perpendicular  to  V.  See  Fig. 
107.  The  circle  whose  center  is  a,  g^ 
represents  the  horizontal  projection 
of  the  base,  and  the  straight  line 
f'-g'  included  between  the  two  ver- 
tical lines/,-/'  and  g-g',  each  drawn 
tangent  to  the  circle  whose  center  is 
ap  represents  the  vertical  projection 
of  the  base.  B  represents  the  vertex 
and  A-B  represents  the  axis. 

The  horizontal  projections  of  the  extreme  elements  B-D  and 
B-JE  may  now  be  determined  as  in  the  previous  case. 

The  lines  b'-f  and  b'-g^  represent  the  vertical  projections  of  the 
highest  and  the  lowest  elements,  and  therefore  the  extreme  elements, 
as  seen  from  the  observer's  standpoint  while  projecting  on  V. 

241.  To  assume  a  Rectilinear  Element  of  the  Cone.  Let  the  cone 
be  represented  with  a  circular  base  on  H,  as  shown  in  Fig;  108. 

Assume  any  point,  as  i>,  on 
the  circumference  of  the  base. 
Connect  D  and  i?  by  a  straight 
line.  B-B,  horizontally  pro- 
jected in  b-df  and  vertically 
projected  in  b'-d\  is  the  required 
j^  element. 

242.  To  assume  a  Point  upon 
the  Surface  of  the  Cone.  Let  the 
cone  be  represented  as  in  Fig. 
108.    Assume    the   horizontal 

Fig.  108  ...  i!  .  i  •    .      x 

projection  o,  of  the  point  at  ran- 
dom. Draw  bf-0f-d,-e,  to  represent  the  horizontal  projection  of 
the  element  containing  the  point  0.    This  element  pierces  H  either 


100  DESCEIPTIVE   GEOMETRY    ' 

at  df  or  at  e,  according  as  0  is  assumed  on  the  upper  or  on  the 
under  surface  of  the  cone. 

Assuming  the  point  on  the  upper  surface  of  the  cone,  the  ele- 
ment in  question  will  be  vertically  projected  at  b'-d',  and  the 
point  0  must  be  vertically  projected  at  o'  on  the  straight  line 
through  0,  perpendicular  to  G-L. 

243.  Shade  Lines.  Shade  lines  upon  cylindrical  and  conical  sur- 
faces, since  such  lines  are  purely  imaginary  elements  of  the  surface, 
unless  they  occur  at  the  intersection  of  base  and  surface,  are  best 
not  represented  by  heavy  lines. 

When  the  plane  of  the  base  of  a  cylinder  or  cone  is  perpendicu- 
lar to  the  plane  of  projection,  the  projection  of  the  base  upon  this 
plane  of  projection  is  a  straight  line,  and  as  a  rule  will  represent 
a  line  in  space,  which  is  partially  a  shade  line  and  partially  not. 

For  the  sake  of  appearance  the  projection  of  the  base  under 
these  conditions  will  be  drawn  either  wholly  a  heavy  line  or 
wholly  a  light  line  according  as  the  portion  which  should  be 
represented  as  a  shade  line  exceeds  or  does  not  exceed  in  length 
that  portion  which  should  be  represented  as  a  light  line. 

In  Fig.  99,  remembering  the  direction  taken  by  the  rays  of 
light,  it  is  evident  that  the  upper  base  of  the  cylinder  and  that 
portion  of  the  surface  to  the  left  and  limited  by  the  two  elements 
D-E  and  F-G  will  be  in  the  light,  and  that  the  remainder  of  the 
surface  will  be  in  the  dark.  Therefore  the  only  shade  lines  to  be 
represented  in  this  case  are  D-M-F  and  E-L-G. 

When  projecting  oniT,  D-M-F  is  visible  and  its  horizontal  projec- 
tion d-m-f,  is  drawn  as  a  heavy  line.  When  projecting  on  Fthe  por- 
tion M-F  of  the  shade  line  D-ilf-i'' and  the  portion  L-G  of  the  shade 
line  E-L-G  are  visible.  Since,  then,  only  the  small  portion  w!-f 
of  the  vertical  projection  of  the  upper  base  should  be  represented 
as  a  heavy  line,  the  whole  line  is  made  a  light  line.  Since  the  larger 
portion  V -g^  of  the  vertical  projection  of  the  lower  base  should  be 
represented  as  a  heavy  line,  the  whole  line  is  made  a  heavy  line. 

In  Fig.  104  the  shade  line  to  be  represented  is  approximately 
D-F-E.  Since  the  larger  portion  f'-e'  of  the  vertical  projection 
of  the  base  should  be  made  a  heavy  line,  the  whole  line  f'-g'  is 
made  a  heavy  line. 


SIKGLE  CURVED  SURFACES  101 

244.  Problem  151.     Given  a  cone  whose  axis  is  oblique  to  If  and 
V  and  whose  base  is  a  circle  on  V;  required  to  draw  the  two  pro- 
jections of  the  cone  and  to  assume  a  point  upon  the  surface. 

245.  Problem  152.  Given  a  cone  whose  axis  is  perpendicular  to 
H,  whose  vertex  is  1  unit  below  If,  and  whose  right  section  at  the 
distance  of  8  units  below  If  is  a  circle  6  units  in  diameter;  required 
to  draw  the  two  projections  of  the  cone  and  to  assume  a  point  upon 
the  surface. 

246.  Problem  153.  Given  a  cone  whose  base  is  a  circle  on  If  and 
whose  axis  is  in  a  profile  plane  and  oblique  to  If;  required  to  draw  the 
two  projections  of  the  cone  and  to  assume  a  point  upon  the  surface. 

247.  The  Convolute.  The  convolute  is  a  single  curved  surface 
which  may  be  generated  by  a  straight  line  moving  tangentially  to 
a  curve  of  double  curvature.  The  generating  line  is  called  the 
generatrix,  the  curved  line  is  called  the  directrix,  and  the  various 
positions  occupied  by  the  generatrix  are  called  the  rectilinear 
elements  of  the  surface. 

Since  a  rectilinear  tangent  to  a  curved  line  contains  two  consecu- 
tive points  of  the  curve,  two  consecutive  tangents  must  have  a 
point  in  common  and  therefore  intersect.  The  consecutive  ele- 
ments of  the  convolute,  then,  will  intersect,  and  since  the  directrix 
is  of  double  curvature,  only  those  elements  which  are  consecutive 
will,  in  general,  intersect. 

If  the  directrix  is  a  helix,  the  convolute  is  called  a  helical 
convolute. 

248.  To  represent  the  Helical  Convolute.  The  helical  convolute 
is  represented  in  Fig.  87,  where  for  purposes  of  clearness  the  dis- 
tances between  consecutive  positions  of  the  generatrix  are  greatly 
magnified. 

The  curve  a -^-C-i>-^-7^ represents  the  helical  directrix.  The 
straight  lines  a^-B-C,  a^^-C-D,  aj^^^-D-E,  etc.,  represent  positions 
of  the  generatrix  or  rectilinear  elements  of  the  surface. 

It  will  be  noticed  here,  as  stated  above,  that  only  consecutive 
elements  intersect;  for  example,  a^^-C-D  intersec^ts  its  preceding 
consecutive  element,  a^-B-C,  at  C;  it  also  intersects  its  succeeding 
consecutive  element,  a^^^-B-E,  at  B ;  but  it  does  not  intersect  the 
element  ay-E-F,  which  is  not  consecutive  to  it. 


102  DESCRIPTIVE   GEOMETRY 

If  we  conceive  a  large  number  of  these  tangents  or  elements  of 
the  surface  to  be  drawn  at  small  distances  apart,  we  may  form  an 
idea  of  the  character  of  the  surface. 

The  curved  line  a,-a,,-a,, -a,,,,-,  etc.,  traced  through  the  various 
points  in  which  the  elements  pierce  H^  is  the  intersection  of  the 
surface  with  H  and  may  be  taken  as  the  base  of  the  surface. 

From  the  method  by  which  these  points  <x„,  a,,,,  a,,,,,  etc.,  are 
found  (see  Section  216),  it  will  be  seen  that  the  base  of  the  helical 
convolute,  in  case  the  axis  of  the  helical  directrix  is  taken  perpen- 
dicular to  H^  is  the  involute  of  that  circle  which  represents  the 
horizontal  projection  of  the  directrix. 

In  Fig.  87  only  that  portion  of  the  surface  extending  between 
the  helical  directrix  and  the  base  on  H  is  considered. 

Since  the  generatrix  extends  without  limit  in  both  directions 
from  the  point  of  tangency  on  the  helix,  there  will  be  generated 
simultaneously  on  opposite  sides  of  the  helical  directrix  two  dis- 
tinct portions  of  the  surface.  These  two  portions  of  the  surface 
are  called  nappes  of  the  surface,  and  their  line  of  separation,  which 
is  the  helical  directrix,  is  called  the  edge  of  regression. 

From  the  nature  of  the  helical  directrix  and  the  relation  of  the 
generatrix  to  the  directrix  it  is  evident  that  if  the  generation  be 
extended  beyond  one  circuit  of  the  axis,  the  surface  will  consist 
of  a  series  of  overlapping  surfaces  whose  base  or  intersection  with 
H  will  have  the  form  of  a  spiral. 

249.  To  assume  a  Rectilinear  Element  of  the  Helical  Convolute. 
Draw  a  rectilinear  tangent  to  the  helical  directrix  (see  Section  216). 

250.  To  assume  a  Point  upon  the  Surface  of  the  Helical  Convolute. 
First  assume  an  element  of  the  surface  and  then  assume  a  point 
upon  the  element. 


CHAPTER  XI 


REPRESENTATION  OF  WARPED  SURFACES 

251.  The  warped  surfac3  may  be  generated  by  a  straight  line  mov- 
ing in  such  a  way  that  its  consecutive  positions  do  not  remain  in  the 
same  plane.  Evidently  there  can  be  as  many  warped  surfaces  as  there 
are  distinct  laws  restricting  the  motion  of  straight  lines  in  this  way. 

252.  The  Warped  Surface  with  Two  Linear  Directrices  and  a  Plane 
Directer.  The  warped  surface  with  two  linear  directrices  and  a 
plane  directer  is  generated  by  the  movement  of  a  straight  line 
which  constantly  touches 
two  linear  directrices  and 
remains  parallel  to  a  plane 
directer. 

253.  The  Hyperbolic 
Paraboloid.  The  hyper- 
bolic paraboloid  is  a  warped 
surface  with  two  rectilin- 
ear directrices  and  a  plane 
directer. 

This  surface  is  repre- 
sented by  locating  the  two 
rectilinear  directrices,  the 
plane  directer,  and  a  suf-  Fig.  109 

ficient  number  of  the  rectilinear  elements,  or  positions  occupied  by 
the  generatrix,  to  reveal  the  character  of  that  portion  of  the  surface 
under  consideration. 

254.  To  assume  a  Rectilinear  Element  of  the  Hyperbolic  Parab- 
oloid. See  Fig.  109.  Let  M-JSf  smd  0-P  represent  the  rectilinear 
directrices,  and  let  *S^  represent  the  plane  directer. 

Assume  any  point,  as  A^  upon  the  directrix  M-N^  and  through  A^ 
by  Section  172,  draw  the  plane  T  parallel  to  .S'.  By  Section  151 
find  the  point  B  in  which  the  directrix  0-P  intersects  S,  and 

103 


104 


DESCRIPTIVE  GEOMETRY 


connect  A  and  ^  by  a  straight  line.  A-B  is  an  element  of  the  sur- 
face, since  it  touches  the  two  directrices  and  lies  in  the  plane  T 
which  is  parallel  to  the  plane  directer. 

255.  To  assume  a  Point  upon  the  Surface  of  the  Hyperbolic 
Paraboloid. 

Analysis  1.  First  assume  an  element  of  the  surface  and  then 
assume  a  point  upon  this  element. 

Analysis  2.  Assume  at  random  the  horizontal  projection  of  the 
point.    Through  the  horizontal  projecting  line  of  the  point  draw 

a  plane  and  find  the  inter- 
section of  this  plane  with 
the  surface.  This  inter- 
section must  contain  the 
point  upon  the  surface, 
and  the  vertical  projection 
of  the  point  must  lie  both 
upon  the  vertical  projec- 
tion of  this  intersection 
and  upon  the  vertical  pro- 
jection of  the  horizontal 
projecting  line  of  the 
point. 

Construction.  See  Fig. 
110.  Let  M-N  and  0-F 
represent  the  two  recti- 
linear directrices,  and  let  N-0  and  M-P  represent  two  rectilinear 
elements  of  the  surface. 

Any  plane  parallel  to  the  two  lines  N-0  and  M-P  is  the  plane 
directer  of  the  surface. 

Assume  at  random  x^  as  the  horizontal  projection  of  the  required 
point.  Through  the  horizontal  projecting  line  of  Xpass  any  plane, 
as  S. 

By  Section  254  assume  a  number  of  rectilinear  elements  of  the 
surface  in  the  vicinity  of  X.  These  elements  are  A-B^  D-E^  F-G, 
K-L^  etc.  These  elements,  beginning  with  A-B^  intersect  the 
plane  *S'  in  the  points  Y,  W,  B,  Q,  etc.  The  line  Y-W-B-Q  is  the 
intersection  of  S  and  the  warped  surface,  and  must  contain  X. 


Fig.  110 


KEPKESENTATION  OF  WARPED  SURFACES    105 


The  point  x\  the  point  in  which  a  straight  line  through  x^  per- 
pendicular to  G-L  intersects  y'-w'-r'-q\  is  the  required  vertical 
projection. 

The  greater  the  number  of  rectilinear  elements  assumed,  the 
more  accurately  will  the  point  x'  be  located. 

256.  The  Warped  Surface  with  two  Curvilinear  Directrices  and  a 
Plane  Directer.  The  warped  surface  with  two  curvilinear  directrices 
and  a  plane  directer  is  generated  by  a  moving  straight  line  which 
always  touches  two  curvilinear  directrices  and  remains  parallel  to 
a  plane  directer. 

257.  To  assume  a  Rectilinear  Element  of  a  Warped  Surface  with 
Two  Curvilinear  Directrices  and  a  Plane  Directer. 

Analysis  and  Construction.  See  Fig.  111.  Let  M—N  and  0-P 
represent  the  two  given  curvilinear  directrices,  and  let  S  represent 
the  given  plane  directer. 

In  the  plane  S  assume  a 
number  of  straight  lines,  as 
E-B,  U-l,  E-2,  E~Z,  etc. 
Through  any  point  A  on 
one  of  the  directrices  M-N 
draw  the  straight  lines  A-L^ 
A-K,  A-G,  A-F,  etc.,  par-  O 
allel  respectively  to  E-B, 
E-1,  E-2,  E-S,  etc.  These 
lines  througli  the  point  A 
lie  in  a  plane  parallel  to  S. 

A-L  intersects  the  hori- 
zontal projecting  surface  of 
0-P  at  L ;  A-K  intersects 
the  same  surface  at  K\ 
A-G  intersects  the  same 
surface  at  G;  and  so  on.  L-K-G-F  is  the  intersection  of  the  plane 
through  the  point  A  with  the  horizontal  projecting  surface  of  0-P, 
The  point  X,  in  which  0-P  crosses  L-K-G-F,  is  the  point  in  which 
0-P  intersects  the  plane  through  the  point  A,  The  straight  line 
A-X  is  the  required  element,  since  it  is  in  a  plane  parallel  to  S  and 
touches  the  two  directrices  M-N  and  0-P, 


Fig.  Ill 


106  DESCRIPTIVE   GEOMETEY 

The  greater  the  number  of  auxiliary  lines  assumed  through 
the  point  A,  the  more  accurately  will  the  line  L-K-G-F  be 
determined. 

258.  To  assume  a  Point  upon  a  Warped  Surface  with  Two  Curvi- 
linear Directrices  and  a  Plane  Directer.    See  Section  255. 

259.  The  Conoid.  The  conoid  is  a  warped  surface  with  a  plane 
directer  and  two  linear  directrices,  one  rectilinear  and  the  other 
curvilinear. 

If  the  rectilinear  directrix  of  a  conoid  is  taken  perpendicular  to 
the  plane  directer,  the  surface  is  called  a  right  conoid.    , 

260.  Problem  154.  Assume  a  rectilinear  element  upon  the  warped 
surface  whose  rectilinear  directrices  are  [^4  =  —  6,  3,  1 ;  B  =  1,  6,  6] 
and  [C  =  4,  —2,  1;  i>  =  4,  —2,  6]  and  whose  plane  directer  is 
^  =  4,  30°,  20°. 

261.  Problem  155.  Assume  a  rectilinear  element  upon  the  warped 
surface  whose  rectilinear  directrices  are  [^  =  —  4,  —  4,  2  ;  ^  =  —  4, 
4,  6]  and  [6'  =  0,  1,  6 ;  D  =  6,  6,  1]  and  whose  plane  directer  is 
^=6,  30°,  60°. 

262.  Problem  156.  Assume  a  rectilinear  element  upon  the  warped 
surface  whose  rectilinear  directrices  aj-e  [A  =  —  6,  —  3,  4 ;  B  =  6, 
—  3,  4]  and  [C  =  —  2,  —  5,  6 ;  i>  =  4, 1,  2]  and  whose  plane  directer 
is  S  =  0,  30°,  60°. 

263.  Problem  157.  Given  a  warped  surface  with  two  curvilinear 
directrices  and  a  plane  directer^  the  latter  coincident  with  V;  required 
to  draw  a  rectilinear  element  of  the  surface. 

264.  Problem  158.  Given  a  warped  surface  with  two  curvilinear 
directrices  and  a  plane  directer,  the  latter  parallel  to  G-L  hut 
oblique  to  H  and  V;  required  to  draw  a  rectilinear  element  of  the 
surface. 

265.  Problem  159.  Given  a  warped  surface  with  one  rectilinear 
directrix  and  one  curvilinear  directrix  and  a  plane  directer;  required 
to  draw  a  rectilinear  element  of  the  surface. 

266.  Problem  160.  Solve  the  preceding  problem  when  the  rectilinear 
directrix  is  perpendicular  to  II  and  the  plane  directer  is  H. 

267.  The  Warped  Surface  with  Three  Linear  Directrices.  The 
warped  surface  with  three  linear  directrices  is  generated  by  a 
straight  line  moving  in  such  a  way  as  to  touch  the  three  directrices. 


REPRESENTATION  OF  WARPED  SURFACES    107 


Under  this  class  of  surfaces  one,  two,  or  all  the  directrices  may 
be  rectilinear,  or  one,  two,  or  all  the  directrices  may  be  curvilinear. 

268.  The  Hyperboloid  of  One  Nappe.  The  hyperboloid  of  one 
nappe  is  a  warped  surface  with  three  rectilinear  directrices. 

This  surface  is  represented  by  locating  the  three  rectilinear 
directrices  and  a  sufficient  number  of  the  rectilinear  elements,  or 
positions  occupied  by  the  generatrix,  to  reveal  the  character  of 
that  portion  of  the  surface  under  consideration. 

269.  To  assume  a  Rectilinear  Element  of  the  Hyperboloid  of  One 
Nappe.  See  Fig.  112.  Let  M-N,  0-P,  and  Q-R  represent  the  three 
rectilinear  directrices. 

Assume  any  point, 
as  A^  upon  the  directrix 
M-N.  Assume  any  two 
points,  as  B  and  Z>, 
upon  the  directrix 
0-P.  Connect  A  and  B, 
and  also  A  and  Z>,  by 
straight  lines  and  pro- 
duce. The  plane  of 
these  two  lines  contains 
the  point  A  and  the 
directrix  0-P.  By  Sec- 
tion 151  find  the  point 
G  in  which  the  remain- 
ing directrix  Q-R  inter- 
sects the  plane  of  the  two 
lines  A-B  and  A-D. 
Connect  A  and  6^  by  a  straight  line  which  intersects  0-P  at  JT, 
since  A-G  and  0-P  are  in  the  same  plane.  A-K-G  is  an  element 
of  the  surface,  since  it  touches  the  three  directrices. 

The  two  points  k^  and  k'  should  be  in  the  same  straight  line 
perpendicular  to  G-L^  which  serves  as  a  check  upon  the  work. 

270.  To  assume  a  Point  upon  the  Surface  of  the  Hyperboloid  of 
One  Nappe.  First  assume  an  element  of  the  surface  and  then 
assume  a  point  upon  the  element,  or  follow  the  directions  of 
Section  255. 


Fig.  112 


108 


DESCEIPTIVE   GEOMETRY 


271.  To  assume  a  Rectilinear  Element  of  a  Warped  Surface  with 
Three  Curvilinear  Directrices. 

Analysis  and  Construction.  See  Fig.  113.  Let  M-JSf^  0-F,  and 
Q-E  represent  the  three  given  curvilinear  directrices. 

Assume  a  point  A  on  one  of  the  directrices  M-N.  Assume  a 
number  of  points,  1,  2,  3,  4,  etc.,  on  one  of  the  other  directrices 
0-F.  Through  A  and  1,  A  and  2,  A  and  3,  etc.,  draw  the  straight 
lines  A-l-B,  A-2-D,  A-^-A\  A-4i-F,  etc.  These  lines  are  elements 
of  a  conical  surface  with  vertex  at  A. 

A-l-B  intersects  the  horizontal  projecting  surface  of  Q-It  at  By 


Fig.  113 


A-2-D  intersects  the  same  surface  at  D;  A-S-F  intersects  the 
same  surface  at  F;  and  so  on. 

B-D-E-F  is  the  intersection  of  the  conical  surface  with  the 
horizontal  projecting  surface  of  Q-R. 

The  point  X  in  which  B-D-E-F  intersects  Q-R  is  the  point  in 
which  Q-R  intersects  the  conical  surface. 

>  Connect  A  and  X  by  a  straight  line.  This  line  must  intersect 
0-F^  since  0-F  lies  upon  the  conical  surface  of  which  A-X  is  an 
element. 

A-Y-X  is    the  required    element,   since   it  touches   the  three 
directrices  Jf-iV,   0-F^  and   Q-R. 


REPRESENTATION  OF  WARPED  SURFACES    109 

The  points  y,  and  y^  should  lie  in  the  same  straight  line  perpen- 
dicular to  G-L^  and  thereby  check  the  work. 

The  greater  the  number  of  auxiliary  lines  drawn  through  the  point 
A^  the  more  accurately  may  the  required  element  be  determined. 

272.  Problem  161.  Assume  a  rectilinear  element  upon  the  hyper- 
holoid  of  one  nappe  whose  three  directrices  are  [^  =  —  6,  6,  2 ; 
B=-l,  -1,  6],  [C  =  -2,  -2,  1;  D-2,  5,  ^,  and  [^=0,  6, 
1;  F=^,  3,  6]. 

273.  Problem  162.  Assume  a  rectilinear  element  upon  the  hyper- 
haloid  of  one  nappe  whose  three  directrices  are  [^  =  —  6,  —  2, 
G  ;'  ^  =  -  1,  5,  2],  [(7  =  0,  2,  2  ;  i>  =  0,  2,  6],  and  [^  =  2,  -  2,  1 ; 
i^-6,  4,  6]. 

274.  Problem  163.  G-iven  a  warped  surface  with  three  curvilinear 
directrices^  one  in  //,  another  in  F,  and  the  third  in  the  third  quadrant; 
required  to  assume  a  rectilinear  element  of  the  surface. 

275.  The  Helicoid.  The  helicoid  is  a  warped  surface  generated 
by  a  straight  line  moving  uniformly  around  and  along  a  rectilinear 
directrix  which  it  intersects  and  with  which  it  makes  a  constant 
angle. 

276.  To  represent  the  Helicoid.  The  helicoid  is  represented  by 
locating  the  directrix  or  axis  (which  is  usually  taken  perpendicular 
to  H)^  a  number  of  the  more  important  rectilinear  elements  of  the 
surface,  and  the  base  or  intersection  of  the  surface  with  H. 

To  do  this  we  must  know  the  angle  which  the  generatrix  makes 
with  the  directrix,  and  the  vertical  distance  through  which  the 
generatrix  moves  for  each  circuit  of  the  directrix. 

In  Fig.  114  let  A-B  represent  the  directrix,  or  axis,  assumed 
perpendicular  to  H.  Through  any  point  C  (<?,,  c^)  on  the  axis 
draw  C-D  parallel  to  V  and  making  the  given  angle  with  the  axis. 
Since  C-D  is  taken  parallel  to  V  its  horizontal  projection  c^-d,  will 
be  parallel  to  G-L^  and  its  vertical  projection  c'-d'  will  make  the 
same  angle  with  a'-V  that  the  generatrix  makes  with  the  directrix. 

Since  the  generatrix  moves  uniformly  around  and  along  the 
directrix,  each  point  of  the  generatrix  will  generate  a  helix  whose 
pitch  will  be  equal  to  the  total  rise  or  fall  of  the  generatrix  per 
revolution,  and  whose  radius  will  be  equal  to  the  distance  of  the 
point  from  the  directrix. 


110 


DESCKIPTIVE   GEOMETRY 


Lay  off  upon  a'-h'  downward  from  c'  the  distance  c '-(?''  equal  to 
the  total  fall  of  the  generatrix  per  revolution,  and  through  e^  draw 
c^-d}^  parallel  to  c'-d' .  The  line  c^'-d^'  is  the  vertical  projection  of 
the  generatrix  at  the  end  of  one  circuit. 

The  point  D  {d,^  d'),  which  is  the  point  in  which  the  generatrix 
in  its  original  position  pierces   H^  will  generate  a  helix  whose 

horizontal  projection  is  the  circle 
d-djj-djjj—djj,f- •  •  • ,  and  whose  ver- 
tical projection  d!-d"-d"^-d""-d^ 
may  be  found  by  Section  214. 

When  the  horizontal  projection 
of  the  point  D  takes  the  position 
c?,,,  or  when  the  horizontal  projec- 
tion of  the  generatrix  takes  the 
position  c-dff,  the  vertical  projec- 
tion, d",  of  D  in  this  position  will 
fall  upon  the  vertical  projection  of 
the  helix  generated  by  I)  and  on 
the  straight  line  through  d^^  per- 
pendicular to  G-L, 

This  shows  that  in  moving  from 
the  first  position  to  this  position 
all  points  of  the  generatrix  have 
moved  downward  a  distance  equal 
to  the  distance  of  d"  below  G-L. 
Therefore,  to  obtain  the  vertical  projection  of  C  for  this  position  of 
the  directrix,  lay  off  upon  a'-h'  downward  from  c'  a  distance  c'-c" 
equal  to  the  distance  of  d"  below  G-L.  The  point  c"  is  the  ver- 
tical projection  sought,  and  the  straight  line  e"-d"  is  the  vertical 
projection  of  the  generatrix  in  this  position.  In  the  same  way  we 
may  find  the  horizontal  and  vertical  projections  of  the  generatrix 
in  other  positions,  as  may  be  seen  from  the  diagram. 

The  generatrix  in  its  first  position  pierces  j^  at  c?,,  in  the  second 
position  it  pierces  II  at  e^^  in  the  third  position  it  pierces  H  at 
/,,  etc.  Since  these  various  positions  of  the  generatrix  represent 
elements  of  the  surface,  the  curve  dj-e-f-g^-  •  •  •  will  represent 
the  base  of  the  surface. 


Fig.  114 


REPRESENTATION  OF  WARPED   SURFAC^ES        111 

If  the  angle  between  the  generatrix  and  the  directrix  is  a  right 
angle,  the  surface  becomes  a  right  conoid.  The  generatrix  is  a 
straight  line  and  moves  in  such  a  way  as  to  touch  two  directrices, 
one  rectilinear  and  the  other  curvilinear,  and  remains  parallel  to  a 
plane  directer  H^  to  which  the  rectilinear  directrix  is  perpendicular. 

The  helicoid  must  not  be  confused  with  the  helical  convolute  in 
which  the  generatrix  moves  tangentially  to  the  helix  and  therefore 
does  not  intersect  the  axis. 

The  helicoid  is  of  practical  interest  since  it  has  a  direct  applica- 
tion in  the  structure  of  the  common  screw  thread,  as  may  be  seen 
in  Sections  282  and  283. 

277.  To  assume  a  Rectilinear  Element  on  the  Surface  of  the 
Helicoid.  Represent  the  generatrix  in  one  of  its  positions  as 
explained  above. 

278.  To  assume  a  Point  upon  the  Surface  of  the  Helicoid.  Follow 
the  method  explained  in  Section  255. 

279.  Problem  164.  Represent  the  helicoid  whose  generatrix  makes 
an  aiigle  of  30  degrees  ivith  the  axis  and  whose  pitch  is  6  units. 

280.  Problem  165.  Represent  the  helicoid  whose  generatrix  makes 
an  angle  q/*  60  degrees  with  the  axis  and  whose  pitch  is  4  units. 

281.  Problem  166.  Represent  the  helicoid  whose  generatrix  makes 
an  angle  of  ^()  degrees  with  the  axis  and  whose  pitch  is  6  units. 

282.  Problem  167.    To  represent  a  triangular-threaded  screw. 

In  Fig.  115  let  A-B  represent  the  axis  of  a  helicoid,  and 
let  D-A  and  D-E  represent  two  generating  elements  of  the  sur- 
face, equally  inclined  to  the  axis,  parallel  to  F,  and  intersecting 
at  D. 

Lay  off  upon  D-A  and  D-E  from  D  the  equal  distances  D-F  and 
D-G^  and  through  F  and  G  draw  the  indefinite  vertical  line  F-K. 

While  the  generating  lines  D-A  and  D-E  generate  their  respec- 
tive helicoidal  surfaces,  the  three  points  F,  D,  and  G  will  generate 
their  respective  helices,  and  the  straight  line  F-K  will  generate  a 
cylindrical  surface  whose  diameter  is  equal  to  l,-fr 

If  in  this  particular  case  the  pitch  of  the  helicoids  is  made  equal 
to  the  distance  F-G^  or  some  multiple  of  it,  the  surface  generated 
by  the  two  sides  D-F  and  D-G  of  the  isosceles  triangle  D-F-G 
will  be  the  surface  of  a  triangular-threaded  screw. 


112 


DESCRIPTIVE  GEOMETRY 


After  a  half  revolution  the  two  generating  lines  will  take  the  posi- 
tions indicated  in  vertical  projection  by  d"-a"  and  d"-e"  respec- 
tively, and  after  a  complete  revolution  they  will  take  the  positions 
indicated  in  vertical  projection  by  d'"-a"'  and  d"'-e"'  respectively. 
We  may  regard  the  thread  of  a  triangular-threaded  screw  as 
generated  by  the  surface  of  an  isosceles  triangle  which  moves  uni- 
formly around  and  along  the  surface  of  a  right  circular  cylinder 
in  such  a  way  that  the  base  of  the  triangle  always  rests  on  the 
surface  of  the  cylinder,  and  the  plane  of  the  triangle  always  contains 

the  axis  of  the  cylinder. 

On  account  of  this  motion  each 
apex  of  the  triangle  will  gener- 
ate a  helix  (see  Section  213). 

If  the  downward  movement 
of  the  triangle  per  revolution 
is  equal  to  its  base,  the  screw  is 
called  single  threaded ;  if  equal 
to  twice  the  base  of  the  triangle, 
it  is  called  double  threaded; 
and  if  equal  to  three  times  the 
base  of  the  triangle,  it  is  called 
triple  threaded. 

Fig.  115  represents  a  single- 
triangular-threaded  screw. 
The  cylinder  is  represented  in 
plan  by  f-n^-l-o^  and  in  eleva- 
tion by  f-y-mJ-V.  The  gen- 
erating triangle,  when  its  plane 
is  parallel  to  F,  is  represented 
in  plan  hj  fj—dj  and  in  eleva- 
tion by  f'-d'-g'. 

The  helices  generated  by  the  points  F,  D,  and  G  may  be  repre- 
sented by  Section  214,  remembering  that  the  pitch  in  each  case  is 
equal  to  the  length  of  the  base  of  the  triangle. 

The  extreme  elements  of  the  thread  surface  on  the  right  and  left 

are  represented  in  vertical  projection  hj  f'-d'-g'-d"'-g"' and 

hy  f"-d''-g"-d"'^-g"" ,  which  represent  the  vertical  projections 


Fig.  115 


REPRESENTATION  OF  WARPED  SURFACES    113 


of  the  sides  of  the  generating  triangle  when  occupying  positions 
parallel  to  V. 

283.  Problem  168.    To  represent  a  square-threaded  screw. 

In  Fig.  116  let  A-B  represent  the  axis  of  a  helicoid,  and  let  D-E 
and  F-G^  equal  in  length,  perpendicular  to  the  axis,  parallel  to 
F,  and  at  a  stated  distance  apart, 
represent  two  generating  ele- 
ments of  the  surface. 

Make  B-K  and  F-L  each 
equal  to  D-F,  Through  D  and 
i^'draw  the  indefinite  vertical 
line  D-M^  and  through  K  and 
L  draw  the  vertical  line  K-N. 

While  the  generating  lines 
D-K  and  F-L  generate  their  G 
respective  helicoidal  surfaces, 
the  four  points  Z>,  F^  i,  and 
K  will  generate  their  respec- 
tive helices,  and  the  two  verti- 
cal lines  D-M  and  A'-iV^  will 
generate  cylindrical  surfaces 
whose  diameters  are  equal  to 
d-d,^  and  k-k,,  respectively. 

If  in  this  particular  case  the 
pitch  of  the  helicoidal  surfaces 
is  made  equal  to  twice,  or  four 
times,  or  six  times,  etc.,  the  distance  D-F^  the  surface  generated 
by  the  three  sides  K-D,  D-F^  and  F-L  of  the  square  K-D-F-L 
will  be  the  surface  of  a  square-threaded  screw. 

In  Fig.  116  the  pitch  is  made  equal  to  four  times  the  distance 
D-F.  After  a  half  revolution  the  three  generating  lines  D-K^ 
D-F^  and  F-L  will  take  the  positions  indicated  in  vertical  projection 
by  d"-k'',  d''-f\  and  f'-l" ;  and  after  a  complete  revolution  they 
will  take  the  positions  indicated  in  vertical  projection  by  d''^-k"'^ 
d'"-r\  and  f"'-l"'. 

We  may  regard  the  thread  of  a  square-threaded  screw  as  gener- 
ated by  the  surface  of  a  square  which  moves  uniformly  around  and 


114  DESCKIPTIVE  GEOMETRY 

along  the  surface  of  a  right  circular  cylinder  in  such  a  way  that 
one  side  of  the  square  always  remains  on  the  surface  of  the  cylinder 
and  the  plane  of  the  square  always  contains  the  axis  of  the  cylinder. 

On  account  of  this  motion  each  apex  of  the  square  will  generate 
a  helix  (see  Section  213). 

If  the  downward  movement  of  the  generating  square  per  revo- 
lution is  equal  to  twice  the  side  of  the  square,  the  thread  is  called 
single  threaded,  if  equal  to  four  times  the  side  of  the  square  it 
is  called  double  threaded,  and  if  equal  to  six  times  tlie  side  of  the 
square  it  is  called  triple  threaded. 

Fig.  116  represents  a  double-square-threaded  screw.  The  inner 
cylinder  is  represented  in  plan  by  k-Oj-k,-p,  and  in  elevation  by 
k' -n' -r' -q' .  The  generating  square  in  its  original  position,  when 
its  plane  is  parallel  to  T,  is  represented  in  plan  by  k-d,  and  in 
elevation  by  k'-d'-f'-V. 

The  helices  generated  by  the  four  vertices  K,  i>,  7'',  and  L  of  the 
square  may  be  represented  by  Section  214,  remembering  that  the 
pitch  of  each  helix  is  equal  to  four  times  the  side  of  the  generating 
square. 

The  thread  directly  below  the  one  just  generated  is  generated 
by  the  square  Z-  W-X-  Y,  which  moves  below  the  square  K-D-F-L 
and  at  a  distance  equal  to  the  side  of  the  square. 

The  spaces  between  the  threads  of  a  square-threaded  screw,  so 
far  as  the  drawing  is  concerned,  may  be  made  equal  in  dimension 
to  the  threads  themselves. 


CHAPTER   XII 

REPRESENTATION  OF   SURFACES   OF  REVOLUTION 

284.  General  Properties  of  Surf  aces  of  Revolution.  The  intersec- 
tion of  a  surface  of  revolution  by  a  plane  perpendicular  to  the 
axis  is  the  circumference  of  a  circle,  since  from  the  nature  of 
the  generation  of  such  surfaces  (see  Section  223)  each  point  of 
the  generatrix  generates  the  circumference  of  a  circle  whose  plane 
is  perpendicular  to  the  axis. 

If  two  surfaces  of  revolution  have  a  common  axis,  and  the  sur- 
faces either  intersect  or  are  tangent  to  each  other,  their  line  of 
intersection  or  of  tangency  will  be  the  circumference  of  a  circle 
whose  center  is  in  the  common  axis  and  whose  plane  is  per- 
pendicular to  the  axis.  For  if  through  the  axis  and  any  point 
of  the  line  of  intersection  or  of  tangency  we  pass  a  plane,  it  will 
cut  from  the  two  surfaces  two  lines  which  will  either  intersect  or 
be  tangent  at  the  assumed  point.  If  now  these  two  lines  with 
their  point  of  intersection  or  of  tangency  be  revolved  about  the 
common  axis,  the  lines  will  generate  their  respective  surfaces, 
and  the  point  of  intersection  or  of  tangency,  which  will  remain 
common  to  the  two  surfaces,  and  therefore  generate  their  line  of 
intersection  or  of  tangency,  will  generate  the  circumference  of  a 
circle  whose  center  is  in  the  axis  and  whose  plane  is  perpendicular 
to  the  axis. 

Two  surfaces  of  revolution  are  tangent  to  each  other  when  they 
are  tangent  to  the  same  surface  at  a  common  point,  or  when  planes 
passed  through  their  point  of  contact  cut  from  the  two  surfaces 
lines  which  are  tangent  to  each  other  at  the  point  of  contact. 

A  plane  tangent  to  a  single  curved  surface  of  revolution  at  a 
given  point  will  be  tangent  to  the  surface  all  along  the  rectilinear 
element  passing  through  this  point  (see  Section  226). 

285.  The  Meridian  Plane  and  the  Meridian  Line.  Any  plane  con- 
taining the  axis  of  a  surface  of  revolution  is  called  a  meridian  plane 
and  its  intersection  with  the  surface  is  called  a  meridian  line. 

115 


116 


DESCRIPTIVE   GEOMETRY 


All  meridian  lines  of  the  same  surface  of  revolution  are  the  same. 
If  a  plane  is  tangent  to  a  surface  of  revolution  at  a  given  point, 
it  will  be  perpendicular  to  the  meridian  plane  of  the  surface  passing 
through  this  point.  For  if  through  the  point  of  tangency  we  pass 
a  plane  perpendicular  to  the  axis,  it  will  cut  from  the  surface  the 
circumference  of  a  circle,  from  the  tangent  plane  a  straight  line  tan- 
gent to  the  circle  at  the  point  of  tangency,  and  from  the  meridian 
plane  a  straight  line  which  is  the  radius  of  the  circle  at  the  point 
of  tangency,  and  therefore  perpendicular  to  the  rectilinear  tangent. 
Through  the  point  of  tangency  and  parallel  to  the  axis  draw  a 
straight  line.  This  line  is  in  the  meridian  plane,  is  perpendicular 
to  the  plane  passed  perpendicular  to  the  axis,  and  is  therefore 
perpendicular  to  the  rectilinear  tangent. 

Since  then  the  tangent  plane  has  a  straight  line  (the  rectilinear 
tangent)  perpendicular  to  two  straight  lines  (the  radius  and  the  line 

parallel  to  the  axis)  of  the  merid- 
ian plane,  it  must  be  perpendicular 
to  the  meridian  plane. 

286.  Representation  of  Surfaces 
of  Revolution.  Surfaces  of  revo- 
lution are  usually  represented  by 
-jr^  assuming  their  axes  perpendicular 
to  H^  although  this  is  not  neces- 
sary. The  intersection  of  the  sur- 
face with  H^  or  the  horizontal  pro- 
jection of  some  important  section 
of  the  surface  made  by  a  plane  per- 
pendicular to  the  axis,  is  taken  for 
the  horizontal  projection  of  the  sur- 
face, and  the  vertical  projection  of 
the  meridian  curve  whose  plane 
is  parallel  to  V  is  taken  as  the  vertical  projection  of  the  surface. 
287.  To  assume  a  Point  upon  any  Surface  of  Revolution. 
Analysis  and  Construction.  Let  the  surface  of  revolution  be 
represented  as  in  Fig.  117,  where  A-B  represents  the  axis,  where 
the  circle  e-f^  represents  the  horizontal  projection  of  the  largest 
horizontal  circle  of  the  surface,  and  where  d'-e'-f'-g'  represents 


SURFACES  OF  REVOLUTION  117 

the  vertical  projection  of  the  meridian  curve  whose  plane  is 
parallel  to  V. 

Assume  at  random  the  horizontal  projection  p^  of  the  required 
point.  The  horizontal  projecting  line  of  the  point  must  intersect 
the  given  surface  in  all  the  points  of  the  surface  which  can  have 
their  horizontal  projection  at  p^ .  Through  this  horizontal  project- 
ing line  and  the  axis  of  the  surface  pass  the  meridian  plane  S, 
which  will  cut  from  the  surface  a  meridian  curve.  The  points  in 
which  this  meridian  curve  is  intersected  by  the  horizontal  project- 
ing line  of  the  point  must  be  the  points  in  question. 

Revolve  the  plane  >S^  about  A-B  as  an  axis  until  the  plane  is 
parallel  to  F.  The  horizontal  trace  will  then  take  the  position  ^S-s,,, 
the  point  p,  will  take  the  position  jo,,,  and  the  vertical  projection 
of  the  horizontal  projecting  line  will  take  the  position  p""-p^-p^^. 
The  plane  of  the  meridian  curve  is  now  parallel  to  V  and  the 
vertical  projection  of  the  meridian  curve  will  be  identical  with  the 
vertical  projection  of  the  surface. 

The  points  p"'\  p^^  p^\  the  points  in  which  the  vertical  projection 
of  the  horizontal  projecting  line  in  revolved  position  intersects  the 
vertical  projection  of  the  meridian  curve  in  revolved  position,  must 
represent  the  vertical  projections  of  the  required  points  in  revolved 
position. 

After  the  counter  revolution  p,j  will  take  the  position  jt?,,  p"^' 
will  take  the  position  p'^  p^  will  take  the  position  p'\  and  p^^  will 
take  the  position  Jt?"^  Therefore  any  one  of  the  three  points  p\  p", 
or  p'"  may  be  taken  as  the  vertical  projection  of  the  point  upon 
the  surface  which  has  its  horizontal  projection  at  p^ . 

The  point  may  be  located  by  assuming  the  vertical  projection 
first. 

288.  Problem  169.  Given  a  cylinder  of  revolution  whose  axis  is 
perpendicular  to  H;  required  to  represent  the  cylinder  and  to  assume 
a  point  upon  the  surface. 

289.  Problem  170.  Given  a  cylinder  of  revolution  whose  axis  is 
parallel  to  G-L;  required  to  represent  the  cylinder  and  to  assume  a 
point  upon  the  surface. 

290.  Problem  171.  Given  a  cone  of  revolution;  required  to  repre- 
sent the  cone  and  to  assume  a  point  upon  the  surface.  , 


118  DESCmPTlVE  GEOMETRY 

291.  Problem  172.  G-ive7i  a  sphere  whose  center  is  in  the  third 
quadrant;  required  to  represent  tJie  sphere  a7id  to  assume  a  point 
upon  the  surface. 

292.  Problem  173.  Given  a  sphere  whose  center  is  in  G-L; 
required  to  represent  the  sphere  and  to  assume  a  point  upon  the 
surface. 

293.  Problem  174.  Given  a  sphere  whose  center  is  in  the  second 
quadrant  and  equidistant  from  H  ajid  V;  required  to  represent  the 
sphere  and  to  assume  a  point  upon  the  siirface. 

294.  Problem  175.  Represent  an  ellipsoid  of  revolution  and  assume 
a  point  upon  the  surface. 

295.  Problem  176.  Represent  a  paraboloid  of  revolutioyi  and  as- 
sume a  point  upon  the  surface. 

296.  Problem  177.  Represent  a  hyperholoid  of  revolution  and  as- 
sume a  point  upon  the  surface. 

297.  The  Hyperboloid  of  Revolution  of  One  Nappe.  The  hyper- 
boloid  of  revolution  of  one  nappe  is  a  warped  surface  of  revolution 
generated  by  the  revolution  of  a  straight  line  about  a  rectilinear 
axis  not  in  the  plane  of  the  generatrix. 

To  represent  the  hyperboloid  of  revolution  of  one  nappe  we  must 
know  the  relation  of  the  generatrix  to  the  axis.  This  may  be 
expressed  by  giving  the  distance  of  the  generatrix  from  the  axis 
and  the  inclination  of  the  generatrix  to  a  plane  perpendicular  to 
the  axis.  The  distance  of  the  generatrix  from  the  axis  will,  of 
course,  be  measured  on  a  straight  line  perpendicular  to  the  two 
(see  Section  199). 

In  Fig.  118  let  ^-7?,  assumed  perpendicular  to  i/,  represent  the 
rectilinear  axis. 

When  the  generatrix  occupies  a  position  parallel  to  F,  the  angle 
which  its  vertical  projection  makes  with  G-L  must  equal  the  angle 
which  the  generatrix  makes  with  H  -,  and  the  perpendicular  dis- 
tance from  the  point  in  which  the  axis  pierces  H  to  the  horizontal 
projection  of  the  generatrix  must  equal  the  distance  of  the  genera- 
trix from  the  axis  (see  Problem  137,  Analysis  2). 

Therefore  through  any  point,  as  e\  on  the  vertical  projection  of 
the  axis  draw  d'-e^-f\  making  w^ith  G-L  the  angle  which  the  gen- 
eratrix is  to  make  with  H.    Draw  d-e-g,  parallel  to  G-L  and  at 


SURFAC  ES  OF  REVOLUTION 


119 


a  distance  from  the  point  in  which  the  axis  pierces  H  equal  to  the 
distance  which  the  generatrix  is  to  be  from  the  axis.  D-E-F 
represents  the  generatrix  in  a  position  parallel  to  V  and  between 
the  axis  and  V. 

From  the  nature  of  the  generation  every  point  of  the  generatrix 
will  generate  the  circumference  of  a  circle  whose  plane  is  perpen- 
dicular to  the  axis  and  whose  radius  is  equal  to  the  distance  of 
the  point  from  the  axis. 

Since,  by  Section  199,  C-E 
is  the  straight  line  perpendic- 
ular to  both  A-B  and  D-E-F, 
the  point  E  of  the  generatrix 
is  nearer  to  the  axis  than  any 
other  point  of  the  generatrix, 
and  will  generate  the  circum-  G 
ference  of  the  smallest  circle 
of  the  surface.  This  circle 
is  horizontally  projected  in 
711,-0 -7i-7n,  and  vertically 
projected  in  m'-n'^  and  is 
called  the  circle  of  the  gorge. 

It  will  be  observed  that  as 
we  increase  the  distance  from 
E,  other  points  upon  the  gen- 
eratrix above  or  below  E  will 
generate  circumferences  of  circles  of  gradually  increasing  radii. 

It  will  be  also  observed  that  the  circumference  of  the  circle 
generated  by  a  point  of  the  generatrix  at  any  given  distance 
above  the  plane  of  the  circle  of  the  gorge  will  have  the  same 
radius  as  the  circumference  of  the  circle  generated  by  a  point  of 
the  generatrix  at  the  same  distance  below  the  plane  of  the  circle 
of  the  gorge. 

This  shows  that  the  surface  is  symmetrical  with  reference  to 
the  plane  of  the  circle  of  the  gorge. 

The  circumference  of  the  circle  d,-p-q,-r,,  generated  by  the 
point  7),  the  point  in  which  the  generatrix  D-E-F  pierces  H,  is 
called  the  base  of  the  surface. 


Fig.  118 


120  DESCRIPTIVE  GEOMETRY 

If  through  E  we  draw  another  straight  line  G-E-K  parallel  to  V 
and  making  the  same  angle  with  H  as  D-E-F^  and  revolve  this 
line  about  the  axis  A-B^  it  will  generate  a  surface  identical  with 
the  one  which  we  have  been  considering,  since  for  any  given  dis- 
tance above  or  below  the  plane  of  the  circle  of  the  gorge  points 
upon  the  two  generatrices  are  equidistant  from  the  axis. 

The  surface  can  then  be  generated  by  the  revolution  of  two  dis- 
tinct straight  lines.  Therefore  through  any  point  of  the  surface 
two  rectilinear  elements  of  the  surface  may  be  drawn. 

Since  every  element  of  the  surface  intersects  the  circumference 
of  the  circle  of  the  gorge,  and  since  the  circle  of  the  gorge  is  the 
smallest  circle  of  the  surface,  the  horizontal  projection  of  each 
element  of  the  surface  must  contain  a  point  in  the  horizontal 
projection  of  the  circumference  of  the  circle  of  the  gorge  and  yet 
cannot  intersect  it.  Therefore  the  horizontal  projection  of  each 
element  of  the  surface  must  be  tangent  to  the  horizontal  projec- 
tion of  the  circle  of  the  gorge. 

298.  To  assume  a  Rectilinear  Element  of  the  Hyperboloid  of  Revo- 
lution of  One  Nappe.  See  Fig.  118.  Assume  any  point,  as  P,  upon 
the  base.  Through  p,  draw  p-o,  tangent  to  the  horizontal  projec- 
tion of  the  circle  of  the  gorge.  This  is  the  horizontal  projection 
of  an  element  of  the  surface  piercing  //  at  P  and  crossing  the 
circle  of  the  gorge  at  0.  P  is  vertically  projected  at  p'  and  0  is 
vertically  projected  at  o'.  Therefore  p'-o'  is  the  vertical  projection 
of  the  element  in  question. 

299.  To  assume  a  Point  upon  the  Surface  of  the  Hyperboloid 
of  Revolution  of  One  Nappe.  First  assume  an  element  of  the 
surface  and  then  assume  a  point  upon  the  element. 

Or  we  may  assume  the  horizontal  projection  of  the  point  at 
random,  through  it  draw  the  horizontal  projection  of  the  element 
containing  the  point,  thence  the  vertical  projection  of  the  same 
element  as  explained  above,  and  finally  the  vertical  projection  of 
the  point  upon  the  vertical  projection  of  the  element. 

300.  To  determine  the  Meridian  Curve  of  the  Hyperboloid  of 
Revolution  of  One  Nappe.  We  shall  consider  the  section  of  the 
surface  made  by  the  meridian  plane  parallel  to  F,  for  the  vertical 
projection  of  this  curve  will  be  equal  in  every  respect  to  the  curve 


SURFACES  OF  REVOLUTION  121 

itself.  In  Fig.  118,  S-s,  is  the  horizontal  trace  of  a  meridian  plane 
parallel  to  V.  This  plane  intersects  the  circumference  of  the 
circle  of  the  gorge  at  M  and  N.  Therefore  m'  and  n'  are  two  points 
in  the  vertical  projection  of  the  required  meridian  curve. 

For  the  same  reason  /  and  q'  are  two  more  points  in  the  vertical 
projection  of  the  curve,  since  B  and  Q  are  the  two  points  in  which 
this  same  meridian  plane  intersects  the  base  of  the  surface. 

To  find  other  points  of  the  curve,  draw  horizontal  planes,  for 
example  T.  This  plane  cuts  the  generatrix  D-E-F  at  X  (x^,  x') 
and  cuts  the  surface  itself  in  the  circumference  of  a  circle  whose 
horizontal  projection  is  x-x^^-Xj,  and  whose  vertical  projection 
is  x"-x^".  This  circumference  intersects  the  meridian  plane  at 
X  (Xff^  x"),  also  at  X  (2-,,^  2;'"),  locating  two  points  in  the  required 
curve. 

Since  the  surface  is  symmetrical  with  reference  to  the  plane  of 
the  circle  of  the  gorge,  the  meridian  curve  in  vertical  projection 
will  be  symmetrical  with  reference  to  the  vertical  projection  of  the 
circle  of  the  gorge.  Therefore,  having  determined  the  points  x'' 
and  x'"  in  the  vertical  projection  of  the  curve,  corresponding  points 
1/"  and  y"'  below  m'-n'  may  be  obtained  by  drawing  U-u'  parallel 
to  m'-n'  and  at  the  same  distance  below  m'-n'  as  T-t'  is  above,  and 
by  taking  1/"  and  y'"  at  the  same  distance  from  a'-b'  as  x"  and  x'". 
By  this  process  we  may  find  as  many  points  in  th«  curve  as  may 
be  desired. 


CHAPTER  XIII 


DETERMINATION  OF  PLANES   TANGENT  TO   SURFACES 
OF   SINGLE   CURVATURE 

301.  General  Instructions.  Note  carefully  the  conditions  imposed 
by  the  problem ;  then  by  an  application  of  the  principles  already 
established  with  reference   to   tangency,  determine  at  least  two 

straight  lines  of  the  required 
plane.  The  points  in  which 
these  Ihies  intersect  H  and 
V  will  be  points  in  the  hori- 
zontal and  vertical  traces  of 
the  required  plane. 

302.  Problem  178.  To 
draw  a  plane  tangent  to  a 
cylinder  at  a  point  on  the 
surface. 

Analysis.  Since  the  sur- 
face is  of  single  curvature, 
the  rectilinear  element  of  the 
surface  through  the  assumed 
point  is  the  element  of  tan- 
gency and  must  be  a  line  of  the  required  plane  (see  Section  226). 
Since  the  required  plane  will  be  tangent  to  the  surface  all  along 
this  element  of  tangency,  a  straight  line  tangent  to  the  surface  at 
any  point  on  this  element  will  be  a  line  of  the  required  plane  (see 
Section  226). 

Case  1.    When  the  base  of  the  cylinder  is  a  circle  on  H. 
Construction,    In   Fig.  119   let  A-B  represent  the  axis  of  the 
cylinder,  and  let  0,  assumed  as  in  Section  234,  represent  the  point 
on  the  surface. 

According  to  analysis  the  element  O-I)  through  0  is  a  line  of 
the  required  plane.    This  element  pierces  //  at  d^  and  pierces  V  at 

122 


Fig.  119 


SUEFACES  OF   SINGLE  CURVATURE 


123 


e\  the  former  a  point  in  the  horizontal  trace  and  the  latter  a  point 
in  the  vertical  trace  of  the  required  plane. 

Through  d,  draw  S-d,-Sf  tangent  to  the  circular  base  at  d.. 
This  is  a  line  of  the  required  plane,  since  it  is  tangent  to  a  curve 
of  the  surface  at  a  point  on  O-I).  As  this  tangent  is  also  in  H  it 
is  the  required  horizontal  trace.  To  draw  the  required  vertical 
trace,  connect  S  with  e'  and  produce. 

Cheek.  Through  F^  any  point  on  the  element  0-7),  draw  F-G 
parallel  to  s-S.  This  must  be  a  line  of  the  required  plane,  since 
it  is  drawn  through  a  point  F  in  the  plane  parallel  to  a  line  s^-S 
also  in  the  plane,  and 
therefore  should  pierce 
V  in  the  vertical  trace 
already  found. 

Case  2.  When  the 
plane  of  the  base  of  the 
cylinder  is  oblique  to  H 
but  perpendicular  to  V. 

Construction.  In  Fig. 
120  let  A-B  represent 
the  axis  of  the  cylinder, 
let  the  circle  with  center 
at  a^  represent  the  hori- 
zontal projection  of  the 
base,  let  the  straight  line 
d'-a'-c'  represent  the 
vertical  projection  of  the  base,  and  let  O  represent  the  point  on 
the  surface. 

The  element  of  tangency  0-D  through  0  pierces  H  at  e,  and 
pierces  V  at  /',  points  in  the  required  traces. 

Through  d,  draw  d^-g^  tangent  to  the  horizontal  projection  of 
the  base.  The  line  d-g,  is  the  horizontal  projection,  and  the  line 
d'-a'-c'  is  the  vertical  projection  of  a  line  tangent  to  the  base  of 
the  cylinder  at  the  point  D  on  the  element  0-D.,  and  therefore  a 
line  of  the  required  plane.  D-G  pierces  H  at  g,  and  pierces  V  at  k'^ 
two  more  points  in  the  required  traces.  The  two  traces  S-s,  and 
S-s'  are  now  located. 


124  DESCRIPTIVE  GEOMETRY 

Check,   Same  as  in  Case  1. 

Case  3.    When  the  cylinder  is  one  of  revolution  with  its  axis  par- 
allel to  G-L. 

Construction.    Let  the  cylinder  be  represented  as  in  Fig.  121. 
.    In  assuming  the  point  upon  the  surface,  take  the  horizontal 
projection  o,  at  random.    Through  this  point  draw  a  profile  plane 
cutting  the  surface  of  the  cylinder  in  the  circumference  of  a  circle 

whose   center   is    C.    The 
point  in  which  this  circum- 
^ ^l\^    / ^ s,      ference   is    intersected   by 


u,         c, 


G- 


B 


'NL  n^  J  ^  h,  the    horizontal    projecting 

_  iv.;^  ^  :  '_  Y^^Q  Qf  Q  xnust  be  the  point 


f"  \  ^^ Li _2/     upon  the  surface.    Revolve 

i  I the  profile  plane  about  its 

"dp  horizontal  trace  as  an  axis 


T 


r 


Fig.  121 


'^ 8'    into  //.    The  circle  C  will 

take   the    position    of   the 

circle  whose  center  is  c^, 
and  the  horizontal  project- 
ing line  will  take  the  posi- 
tion 0-0 jj.  Therefore  either  Ojj  or  hjj  may  be  taken  as  the  revolved 
position  of  the  point  upon  the  surface.  Let  us  take  the  point  0, 
which  in  this  revolved  position  is  vertically  projected  at  o^  but  in 
true  position  is  vertically  projected  at  o'. 

The  element  0-D  through  0  is  a  line  of  the  required  plane,  and 
since  this  line  is  parallel  to  G-L^  we  know  that  the  traces  of  the 
required  plane  will  both  be  parallel  to  G-L, 

Through  Ojj  draw  e-Ojj-f^j  tangent  at  Oj^  to  the  circle  whose 
center  is  Cjj, 

This  line  is  the  revolved  position  of  a  line  tangent  to  a  curve 
of  the  surface  at  a  point  on  0-D,  and  therefore  is  the  revolved 
position  of  a  line  of  the  required  plane.  The  line  JE-F  in  true 
position  pierces  H  at  e^  a  point  in  the  required  horizontal  trace, 
and  pierces  V  in  f\  a  point  in  the  required  vertical  trace.  The 
two  traces  S-s^  and  S-s'  are  now  located. 

Chech.  Pass  another  profile  plane  cutting  from  the  surface  of 
the  cylinder  the  circumference  of  a  circle,  and  from  0-D  a  point 


SURFACES  OF  SmGLE  CURVATURE  125 

upon  this  circumference.  The  rectilinear  tangent  to  the  circle  at 
this  point  is  a  line  of  the  required  plane  and  should  pierce  H  and  V 
in  the  traces  already  found. 

303.  Problem  179.  Given  a  cylinder  whose  base  is  a  circle  on  V 
and  whose  axis  is  oblique  to  H  and  V ;  required  to  assume  a  point 
upon  the  surface  and  to  draw  a  plane  tangent  to  the  surface  at  this 
point. 

304.  Problem  180.  Given  a  cylinder  whose  axis  is  oblique  to  II 
and  F,  the  plane  of  whose  base  is  perpendicular  to  H  but  oblique  to  F, 
and  the  vertical  projection  of  whose  base  is  a  circle;  required  to 
assume  a  point  upon  the  surface  and  to  draw  a  plane  tangent  to  the 
surface  at  this  point. 

305.  Problem  181.  Given  a  cylinder  whose  axis  is  [^  =  0,  4,  0 ; 
^  =  0,  8,  6]  and  whose  base  on  H  is  a  circle  of  S-unit  radius ; 
required  to  assume  a  point  upon  the  surface  and  to  draw  a  plane 
tangent  to  the  surface  at  this  point. 

306.  Problem  182.  Given  a  cylinder  whose  axis  is  oblique  to  H  and 
V  and  whose  base  is  a  circle  in  a  plane  parallel  to  H  and  above  it ; 
required  to  assume  a  point  upon  the  surface  and  to  draw  a  plane 
tangent  to  the  surface  at  this  point. 

307.  Problem  183.  To  draw  a  plane  tangent  to  a  cylinder  and 
through  a  point  without  the  surface. 

Analysis.  The  required  plane  will  contain  an  element  of  the 
surface  of  the  cylinder.  Therefore  a  straight  line  through  the 
given  point  and  parallel  to  the  elements  of  the  cylinder  will  be 
a  line  of  the  required  plane  (see  Section  42)  and  should  pierce  H 
and  V  in  points  of  the  required  traces. 

Since  the  required  plane  will  be  tangent  to  the  surface  all  along 
the  element  of  tangency,  any  plane  oblique  to  the  elements  of  the 
cylinder  will  cut  from  the  surface  a  curved  line,  from  the  auxiliary 
line  parallel  to  the  elements  a  point,  and  from  the  required  plane 
a  straight  line  passing  through  this  point  and  tangent  to  the  curve 
at  a  point  on  the  element  of  tangency.  Therefore,  if  through 
any  point  of  the  auxiliary  line  we  draw  a  plane  cutting  from  the 
surface  of  the  cylinder  a  curved  line,  the  straight  line  through 
the  assumed  point  and  tangent  to  the  curve  will  be  a  line  of  the 
required  plane. 


126 


DESCRIPTIVE  GEOMETRY 


Case  1.    When  the  base  of  the  cylinder  is  a  circle  on  H. 

Construction.  In  Fig.  122  let  A-B  represent  the  axis  of  the 
cylinder,  and  let  0  represent  the  point  assumed  without  the  surface. 

Through  0  draw  0-1)  parallel  to  the  elements  of  the  cylinder 
and  produce  it  to  pierce  //  at  d^ .  Through  d^  draw  s-d-e-S  tan- 
gent to  the  circular  base  at  e, .  According  to  analysis  this  is  a  line 
of  the  required  plane,  and  since  it  is  in  H^  it  is  the  required  hori- 
zontal trace.  Through  0  draw  0-G  parallel  to  s-S  piercing  F  at  ^' . 
S-g'-s'  is  the  required  vertical  trace. 

The  element  E-F  through  U  is  the  element  of  tangency. 

Since  from  the  point  c?,  another  straight  line  may  be  drawn  tan- 
gent to  the  base  of  the  cylinder,  another  plane  answering  the  con- 
ditions of  the  problem  may  be  constructed. 

Check.  Note  whether  the  element  of  tangency,  which  is  a  line 
of  the  required  plane,  pierces  V  in  the  vertical  trace  already  located. 


Fig.  122 


Case  2.  When  the  plane  of  the  base  of  the  cylinder  is  perpen- 
dicular to  H  but  oblique  to  V. 

Construction.  In  Fig.  123  let  A-B  represent  the  axis  of  the 
cylinder,  and  let  0  represent  the  point  without  the  surface. 

Through  0  draw  0-D  parallel  to  the  elements  of  the  cylinder 
and  produce  it  to  pierce  H  at  d,  and  to  pierce  V  at  e'. 


SURFACES  OF  SINGLE  CURVATURE 


127 


Through  F^  the  point  in  which  0-D  pierces  the  plane  of  the 
base  of  the  cylinder,  draw  a  rectilinear  tangent  to  the  curve  of 
the  base.  This  tangent  will  be  vertically  projected  in  f-g\  tan- 
gent to  the  vertical  projection  of  the  base,  and  horizontally  pro- 
.jected  in  f-g -a f.  F-G  is  by  analysis  a  line  of  the  required  plane 
and  pierces  H  at  I,  and  pierces  V  at  m'.    S-s,  drawn  through  c?,  and  l^ 


Fig.  123  . 

is  the  required  horizontal  trace,  and  S-s'  drawn  through  e'  and  m' 
is  the  required  vertical  trace. 

The  element  of  tangency  is  that  element  determined  by  the 
point  G. 

Since  another  straight  line  may  be  drawn  through  F  tangent  to 
the  curve  of  the  base,  another  plane  answering  the  conditions  of  the 
problem  may  be  constructed. 

Check.  Note  whether  the  element  of  tangency,  which  is  a  line 
of  the  required  plane,  pierces  //and  Fin  the  traces  now  located. 

Case  3.  When  the  cylinder  is  one  of  revolution  with  its  axis  paral- 
lel to  G-L. 


128 


DESCEIPTIVE  GEOMETRY 


Construction.  In  Fig.  124  let  A-B  represent  the  axis  of  the 
cylinder,  and  let  O  represent  the  point  without  the  surface. 

A  straight  line  through  0  parallel  to  the  axis  A-B  will  be  a  line 
of  the  required  plane,  and  since  this  line  is  parallel  to  G-L^  we 

know  that  both  traces  of 
the  required  plane  will  be 
parallel  to  G-L. 

Through  0  pass  a  profile 
plane  cutting  the  cylinder 
in  the  circle  whose  center 
is  C.  Revolve  this  profile 
plane  about  its  horizontal 
trace  into  H.  O  will  fall 
at  Ojj  and  the  circle  C  will 
take  the  position  of  the 
circle  whose  center  is  Cjj. 
Through  Ojj  draw  Ojj-d^  tan- 
gent to  the  circle  c^.  Ac- 
cording to  analysis  this  is 
the  revolved  position  of  a 
line  of  the  required  plane. 
In  true  position  this  line 
pierces  ^  at  6?,,  a  point  in  the  required  horizontal  trace ;  and 
pierces  V  at  e\  a  point  in  the  required  vertical  trace.  The  two 
traces  S-s,  and  S-s'  are  now  located. 

The  element  of  tangency  is  that  one  determined  by  the  point  K, 
Since  through  o^  another  straight  line  may  be  drawn  tangent 
to  the  circle  Cjj,  another  plane  answering  the  conditions  of  the 
problem  may  be  constructed. 

Check.  Through  some  point  of  the  element  of  tangency  draw  a 
profile  plane  cutting  from  the  cylinder  a  circle.  A  rectilinear  tan- 
gent to  this  circle  at  the  point  assumed  is  a  line  of  the  required 
plane  and  should  pierce  H  and  V  in  the  traces  already  located. 

308.  Problem  184.  Given  a  cylinder  in  the  third  quadrant  whose 
axis  is  oblique  to  H  and  V  and  whose  base  is  a  circle  on  V;  required 
to  assume  a  point  without  the  surface  and  to  draw  through  this  point 
a  plane  tangent  to  the  cylinder. 


s 

d, 

\ 
\ 
\ 
\ 

S, 

"' 

k,  /-       A/.V 

a. 

1        h 

c, 

\b. 

0, 

i    \     \ 

ry 

i r-'--p\ 

J> 

)      ;           i;^ 

'      /         //     i 

a' 

b' 

8 

o 

t"-? 

s' 

e' 

Fig.  124 

SURFACES  OF   SINGLE  CURVATURE 


129 


309.  Problem  185.    (xive7i  a  cyliiider  whose  axis  is  oblique  to  H  and 

F,  the  plane  of  whose  base  is  perpendicular  to  V  but  oblique  to  H^ 

and  the  horizontal  projection  of  whose  base  is  a  circle;  required  to 

assume  a  point  without  the  surface  and  to  draw  a  plane  through  this 

point  tangent  to  the  cylinder. 

310.  Problem  186.  Given  a  cylinder  whose  axis  is  oblique  to  H 
and  V  and  whose  base  is  a  circle  in  a  plane  parallel  to  V  and  in 
front  of  it ;  required  to  assume  a  point  without  the  surface  and  to 
draw  through  this  point  a  plane  tangent  to  the  cylinder. 

311.  Problem  187.  Given  a  cylinder  whose  axis  is  [^  =  —  4,  0, 
—  4  ;  ^  =  —  4,  6,  4]  and  whose  base  on  V  is  a  circle  of  S-unit  radius; 
also  given  the  point  0  =  2,  4,  2  ;  required  to  draw  through  the  point 
a  plane  tangent  to  the  cylinder. 

312.  Problem  188.  To  draw  a  plane  tangent  to  a  cylinder  and 
parallel  to  a  given  straight  line. 

Analysis.  Since  the  required  plane  must  contain  an  element  of 
the  cylinder  and  must  be  parallel  to  the  given  line,  the  required 
plane  will  be  parallel  to 
any  plane  which  is  paral- 
lel both  to  the  axis  of  the 
cylinder  and  to  the  given 
line.  Therefore,  through 
the  given  line  pass  a  plane 
parallel  to  the  axis  of  the 
cylinder  and  draw  the  re- 
quired plane  tangent  to 
the  cylinder  and  parallel 
to  this  plane. 

Case  1.  When  the  base 
of  the  cylinder  is  a  circle 
on  H. 

Construction.  In  Fig. 
125  let  A-B  represent 
the  axis  of  the  cylinder  and  let  M-N  represent  the  given  line. 

Through  any  point  D  of  M-N  draw  D-E  parallel  to  A-B. 
D-E  pierces  H  at  e,  and  pierces  F  at/';  M-N  pierces  H  at  g,  and 
pierces   Fat  k^.    The  plane    U  whose  traces  are  located  by  the 


Fig.  125 


130 


DESCRIPTIVE  GEOMETRY 


points  just  found  will,  by  analysis,  be  parallel  to  the  required  plane. 
The  horizontal  trace  of  the  required  plane,  then,  must  be  parallel 
to  U-u,  and  must  also  be  tangent  to  the  circular  base  on  H  whose 
center  is  a^ .  Therefore  draw  the  horizontal  trace  S-s^  under  these 
conditions,  and  through  S  draw  the  vertical  trace  S-s'  parallel  to 

The  element  of  tangency  may  be  located  by  drawing  through  Z, 
the  point  of  tangency  on  the  base,  a  straight  line  parallel  to  the 
axis. 

Since  another  straight  line  can  be  drawn  parallel  to  J^-u,  and 
tangent  to  the  circle  whose  center  is  «^,  another  plane  answering 

the  conditions  of  the  prob- 
/  lem  may  be  constructed. 

Chech,  Through  some 
point  on  the  element  of 
tangency  draw  a  straight 
line  parallel  to  M-N  and 
note  whether  it  pierces  H 
and  V  in  the  traces 
already  located. 

Case  2.  Wh  e7i  the 
plane  of  the  base  of  the 
cylinder  is  perpendicular 
to  V  hut  ohllque  to  H, 

Construction.  In  Fig. 
126  let  A-B  represent  the 
axis  of  the  cylinder  and 
let  M-N  represent  the 
given  line. 

Through  any  point  D 
of  the  given  line  M-N  draw  D-E  parallel  to  the  axis  A-B.  The 
plane  of  the  two  lines  M-N  and  D-E  is  parallel  to  the  required 
plane.  D-E  pierces  the  plane  of  the  base  of  the  cylinder  at  E^ 
and  M-N  pierces  the  same  plane  at  F.  The  line  E-F  is  the  inter- 
section of  the  auxiliary  plane  with  the  plane  of  the  base  of  the 
cylinder,  and  must  be  parallel  to  the  intersection  of  the  required 
plane  with  the  plane  of  the  base. 


Fig.  126 


SURFACES  OF   SINGLE  CURVATURE 


131 


The  intersection  of  the  required  plane  with  the  plane  of  the 
base  must  also  be  tangent  to  the  curve  of  the  base.  Therefore 
k-g-li^  drawn  parallel  to  e-f^  and  tangent  to  the  circle  whose 
center  is  a^  is  the  horizontal  projection,  and  k'-g'-V  is  the  vertical 
projection  of  the  intersection  of  the  required  plane  with  the  plane 
of  the  base.  K-G-L,  which  is  a  line  of  the  required  plane,  inter- 
sects //  at  kf  and  intersects  V  at  l\  points  respectively  in  the  hori- 
zontal and  vertical  traces  of  the  required  plane. 

Through  G  draw  the  element  of  tangency  G-0,  another  line  of 
the  required  plane,  and  produce  it  to  pierce  H  at  o,  and  to  pierce 
V^tp'.  The  two  required 
traces  S-o^-k-s,  and  >S'-^'- 
p'-s'  may  now  be  drawn. 

Since  another  straight 
line  can  be  drawn  parallel 
to  e-fj  and  tangent  to  the 
circle  whose  center  is  a,, 
another  plane  answering 
the  conditions  of  the  prob- 
lem may  be  constructed. 

Check.  Through  some 
point  of  the  element  of 
tangency  draw  a  straight 
line  parallel  to  the  given 
line  M-X^  and  note 
whether  it  intersects  H  and  V  in  the  traces  now  located. 

Case  3.  When  the  cylinder  is  one  of  revolution  with  its  axis  par- 
allel to  G-L. 

Construction.  In  Fig.  127  let  A-B  represent  the  axis  of  the 
cylinder  and  let  M-N  represent  the  given  line. 

Since  the  axis  of  the  cylinder  is  parallel  to  G-L^  the  traces  of 
the  required  plane  will  also  be  parallel  to  G-L. 

Through  M-N  draw  the  plane  U  parallel  to  the  axis  A-B.  This 
auxiliary  plane  must  be  parallel  to  the  required  plane. 

Draw  a  profile  plane  P  cutting  the  cylinder  in  the  circle 
whose  center  is  C,  and  cutting  the  auxiliary  plane  U  in  the  line 
F-G.     The    intersection    of   the    required    plane    by    the    profile 


132  DESCEIPTIVE  GEOMETEY 

plane  must  be  parallel  to  F-G  and  be  tangent  to  the  circle  C,  since 
the  required  plane  is  both  parallel  to  the  auxiliary  plane  and  tan- 
gent to  the  surface  of  the  cylinder  (see  Section  226). 

Revolve  the  profile  plane  about  its  horizontal  trace  into  H.  The 
circle  C  falls  at  the  circle  whose  center  is  c^,  and  the  line  F-G 
falls  2itf-gjj. 

The  line  k-ljj  drawn  parallel  to  ^,-^^  and  tangent  to  the  circle 
Cjj  is  the  revolved  position  of  the  intersection  of  the  required  plane 
with  the  profile  plane.  In  true  position  K-L  pierces  H  at  k,  and 
pierces  V  at  V,  points  respectively  in  the  horizontal  and  vertical 
traces  of  the  required  plane.  The  two  traces  S-s^  and  S-s'  may 
now  be  drawn. 

The  element  of  tangency  may  be  determined  by  drawing  through 
i?,  the  point  of  tangency  on  the  circle  C,  a  straight  line  parallel 
to  the  axis  A-B. 

Since  another  straight  line  tangent  to  the  circle  Cjj  and  parallel 
to  f-Qji  may  be  drawn,  another  plane  answering  the  conditions 
of  the  problem  may  be  constructed. 

Check.  Through  some  point  on  the  element  of  tangency  draw 
a  straight  line  parallel  to  the  given  line  M-N^  and  note  whether 
this  line  pierces  H  and  V  in  the  traces  now  located. 

313.  Problem  189.  Given  a  cylinder  whose  axis  is  oblique  to  H 
and  V  and  whose  base  is  a  circle  on  V ;  required  to  assume  a  straight 
line  and  to  draw  a  plane  tangent  to  the  cylinder  and  parallel  to 
the  line. 

314.  Problem  190.  Given  a  cylinder  whose  axis  is  oblique  to  H 
and  F,  the  plane  of  whose  base  is  perpendicular  to  H  but  oblique  to 
F",  and  the  vertical  projection  of  whose  base  is  a  circle ;  required  to 
assume  a  straight  line  and  to  draw  a  plane  tangent  to  the  cylinder 
and  parallel  to  the  line. 

315.  Problem  191.  Given  a  cylinder  whose  axis  is  parallel  to  G—L; 
required  to  assume  a  straight  line  in  a  profile  plane  a7id  to  draw 
a  plane  tangent  to  the  cylinder  and  parallel  to  the  line. 

316.  Problem  192.  Given  a  cylinder  whose  axis  is  oblique  to  H 
and  V  and  whose  base  is  a  circle  in  a  plane  parallel  to  H  and  above 
it;  required  to  assume  a  straight  line  parallel  to  G-L  and  to  draw 
a  plane  tangent  to  the  cylinder  and  parallel  to  the  line. 


SURFACES   OF  SINGLE   CURVATURE 


13; 


317.  Problem  193.  To  draw  a  plmie  tangent  to  a  cone  at  a  point 
on  the  surface. 

Analysis.  Since  the  surface  of  a  cone  is  of  single  curvature,  the 
rectilinear  element  of  the  surface  through  the  point  of  tangency  is 
the  element  of  tangency,  and  is  therefore  a  line  of  the  required 
plane  (see  Section  226). 

-  Since  the  required  plane  will  be  tangent  to  the  surface  all  along 
this  element,  a  line  tangent  to  the  surface  at  any  point  on  this 
element  will  be  a  line  of  the  required  plane. 

Case  1.    When  the  base  of  the  cone  is  a  circle  on  H. 

Construction.  In  Fig.  128  let  A-B  represent  the  axis  of  the  cone, 
and  let  0,  assumed  as  in  Section  242,  represent  the  point  on  the 
surface. 

The  element  B-O-D,  which  is  a  line  of  the  required  plane,  pierces 
H  at  df  and  pierces  V  at  e',  points  in  the  required  traces.    Through 


Fig.  128 


df  draw  S-d^-s^  tangent  to  the  circular  base  at  d, .  This  is  a  line  of 
the  required  plane,  because  it  is  tangent  to  a  curve  of  the  surface 
at  a  point  on  0-D.  Since  this  tangent  is  also  in  H,  it  is  the  required 
horizontal  trace.  Connect  S  and  e'  by  a  straight  line  and  produce 
it  to  represent  the  required  vertical  trace. 

Check.  Through  F,  any  point  on  the  element  of  tangency  0-D, 
draw  F-G  parallel  to  s^-S.  This  line  must  be  a  line  of  the  required 
plane,  since  it  is  drawn  through  a  point  F  in  this  plane  and  parallel 
to  a  line  s^-S  also  in  this  plane,  and  should  therefore  pierce  V  in 
the  vertical  trace  already  found. 


134 


descriptivp:  geometry 


Case  2.  Whe^i  the  plane  of  the  base  of  the  cone  in  perpendicular 
to  V  hut  oblique  to  H. 

Construction.  In  Fig.  129  let  A-B  represent  the  axis  of  the  cone, 
let  the  circle  whose  center  is  a^  represent  the  horizontal  projection 
of  the  base,  let  the  straight  line  g'-k'-a'-d'  represent  the  vertical 
projection  of  the  plane  of  the  base,  and  let  0  represent  the  point 
on  the  surface. 

The  element  B-O-D  through  0  pierces  II  at  e,  and  pierces  Fat 
/',  points  in  the  required  traces. 

Through  d,  draw  d-g,  tangent  to  the  horizontal  projection  of 
the  base.  The  line  d-g,  is  the  horizontal  projection  and  d'-g'  is 
the  vertical  projection  of  a  line  tangent  to  the  base  of  the  cone  at 
a  point  D  on  the  element  B-O-D,  and  therefore  a  line  of  the  required 

plane.  D-G  pierces  If  at  g^  and 
pierces  V  at  k',  two  more  points 
in  the  required  traces. 

Check.    Same  as  in  Case  1. 

318.  Problem  194.  Given  a 
cone  whose  axis  is  oblique  to  H 
and  V  and  whose  base  is  a  circle 
on  V ;  required  to  drau)  a  plane 
tangent  to  the  cone  at  a  point  on 
the  surface. 

319.  Problem  195.  G-iven  a 
co7ie  whose  axis  is  oblique  to  H 
and  V  and  whose  base  is  iii  a 
plane  perpendicular  to  H  but 
oblique  to  V ;  required  to  draw 

a  plane  tangent  to  the  cone  at  a  point  on  the  surface. 

320.  Problem  196.  Given  a  cone  whose  axis  is  oblique  to  H  and 
V  and  whose  base  is  a  circle  in  a  plane  parallel  to  H  and  above 
it ;  required  to  draw  a  plane  tangent  to  the  cone  at  a  point  on  the 
surface. 

321.  Problem  197.  To  draw  a  plane  tangent  to  a  cone  and  through 
a  point  without  the  surface. 

Analysis.  Since  the  required  plane  must  contain  the  vertex  of 
the  cone,  a  straight  line  through  the  vertex  of  the  cone  and  the 


SUKFACE8  OF   SINGLE  CURVATURE 


135 


given  point  must  be  a  line  of  the  required  plane.  The  remainder 
of  the  analysis  differs  little  from  that  given  in  connection  with 
Problem  183,  Section  307. 

Case  1.    When  the  base  of  the  cone  is  a  circle  on  If. 

Construction.  In  Fig.  130  let  A-B  represent  the  axis  of  the 
cone,  and  let  0  represent  the  point  without  the.  surface. 

Connect  0  and  B  by  a  straight  line  and  produce  it  to  pierce  H 
at  df  and  to  pierce  F  at  e'.  Through  d^  draw  d^-S-f-s^  tangent  to 
the  circular  base  at  /, . 

According  to  analysis  this  is  a  line  of  the  required  plane,  and 
since  it  is  in  iT,  it  is  the  required  horizontal  trace.  Through  S  and 
e'  draw  the  required  vertical  trace  S—s'. 


Fig.  130 

The  element  B-F  is  the  element  of  tangency  and  a  line  of  the 
required  plane. 

Since  from  the  point  d^  another  straight  line  may  be  drawn  tan- 
gent to  the  base,  another  plane  answering  the  conditions  of  the 
problem  may  be  constructed. 

Check.  Note  whether  the  element  of  tangency  pierces  V  in  the 
vertical  trace  now  located. 

Case  2.  When  the  plane  of  the  base  of  the  cone  is  perpendicular 
to  II  hut  oblique  to  }'. 

Construction.  In  Fig.  131  let  A-B  represent  the  axis  of  the 
cone  and  let  0  represent  the  point  without  the  surface. 


136 


DESCRIPTIVE  GEOMETRY 


Connect  the  vertex  of  the  cone  and  0  by  a  straight  line  and 
produce  it  to  pierce  H  in  d^  and  to  pierce  V  in  e\  points  in 
the  required  traces. 

Through  F,  the  point  in  which  B-0  intersects  the  plane  of  the 
base  of  the  cone,  draw  a  rectilinear  tangent  to  the  curve  of  the 
base.    This  tangent  will  be  vertically  projected  in  f-g'^  tangent 

to  the  vertical  projection  of 
the  base,  and  will  be  horizon- 
tally projected  inf-g-a,. 

By  analysis  7*^-6^  is  a  line 
of  the  required  plane  and 
pierces  H  at  k^  and  pierces 
F"  at  v.  S-Sf  drawn  through 
df  and  k^  is  the  required  hori- 
zontal trace,  and  S-s'  drawn 
through  e'  and  V  is  the  re- 
quired vertical  trace. 

The  element  of  tangency 
is  the  straight  line  through  G 
and  the  vertex  of  the  cone. 
Since  another  straight 
line  may  be  drawn  through 
F  tangent  to  the  curve  of  the  base,  another  plane  answering  the 
conditions  of  the  problem  may  be  constructed. 

Check.  Note  whether  the  element  of  tangency,  which  is  a  line 
of  the  required  plane,  pierces  If  and  V  in  the  traces  now  located. 

322.  Problem  198.  Given  a  cone  whose  axis  is  oblique  to  H  and  V 
and  whose  base  is  a  circle  on  V ;  required  to  assume  a  point  tvith- 
out  the  surface  and  to  draw  through  this  point  a  plane  tangefit  to 
the  cone. 

323.  Probem  199.  Given  a  cone  whose  axis  is  oblique  to  H  and  V, 
the  plane  of  whose  base  is  perpendicular  to  V  but  oblique  to  H,  and 
the  horizontal  projection  of  whose  base  is  a  circle  ;  required  to  assume 
a  point  without  the  surface  and  to  draw  a  plane  through  this  point 
and  tangent  to  the  cone. 

324.  Problem  200.  Given  a  cone  whose  axis  is  oblique  to  H  and  F, 
and  whose  base  is  a  circle  in  a  plane  parallel  to  II  and  above  it ; 


Fig.  131 


SUKFACES  OF   SINGLE  CUEVATUKE 


137 


required  to  assume  a  poirit  without  the  surface  and  to  draw  a  plane 
through  this  point  and  tangent  to  the  cone. 

325.  Problem  201.  To  draw  a  plane  tangent  to  a  cone  and  parallel 
to  a  given  straight  line. 

Analysis.  Since  the  required  plane  must  contain  the  vertex  of 
the  cone,  a  straight  line  through  the  vertex  of  the  cone  and  par- 
allel to  the  given  line  will  be  a  line  of  the  required  plane.  A  plane 
containing  this  auxiliary  line  and  tangent  to  the  cone  will  be  the 
required  plane. 

Case  1.    When  the  base  of  the  cone  is  a  circle  on  IT. 

Construction.  In  Fig.  132  let  A-B  represent  the  axis  of  the  cone 
and  let  M-N  represent  the  given  line. 

Through  the  vertex  of  the  cone  draw  B-D  parallel  to  M-N, 
piercing  //  at  c?,  and  piercing  V  at  e'.    Through  d^  and  tangent  to 


Fig.  132 


the  circular  base  draw  the  required  horizontal  trace  d-S-s^. 
Through  S  and  e'  draw  the  required  vertical  trace  S-e'-s'. 

The  element  of  tangency  is  the  straight  line  drawn  through  the 
vertex  of  the  cone  and  the  point  of  tangency  F  on  the  base. 

Since  another  straight  line  may  be  drawn  through  d,  and  tan- 
gent to  the  circular  base,  another  plane  answering  the  conditions 
of  the  problem  may  be  constructed. 

Check.  Note  whether  the  element  of  tangency  pierces  V  in  the 
vertical  trace  already  located. 


138 


DESCKIPTIVE   GEOMETKY 


Case  2.    When  the  plane  of  the  base  of  the  cone  is  perpefidicular 

to  V  but  oblique  to  H. 

Construction.    In   Fig.  133   let  A-B  represent  the   axis  of   the 

cone  and  let  M-N  represent  the  given  line. 

Through  the  vertex  of  the  cone  draw  E-B-D  parallel  to  IT-iV, 

piercing  H  at  d,  and  piercing  V  at  e\  points  in  the  required  traces. 

Through  F^  the  point  in  which  B-D  intersects  the  plane  of  the 

base  of  the  cone,  draw  G- 
F-0  tangent  to  the  curve 
of  the  base.  The  horizon- 
tal projection  of  this  tan- 
gent is  g^-ff-Of^  and  the 
vertical  projection  of  this 
tangent  is  g'-f'-o'.  By 
analysis  G—F-0  is  a  line 
of  the  required  plane, 
and  pierces  //  at  g^  and 
pierces  V  at  k! .  Through 
g^  and  d^  draw  the  required 
horizontal  trace  S-s^. 
Through  k'  and  e'  draw 
the  required  vertical 
trace  S-s'. 

The   element  of  tan- 

gency  is  the  straight  line  drawn  througli  the  point  of  tangency  on 

the  base  and  the  vertex  of  the  cone. 

Since  another  straight  line  may  be  drawn  through  F  and  tangent 

to  the  curve  of  the  base,  another  plane  answering  the  conditions 

of  the  problem  may  be  constructed. 

Check.    Note  whether  the  element  of  tangency  intersects  //  and 

V  in  the  traces  now  located. 

326.  Problem  202 .  G-iven  a  cone  whose  aMS  is  oblique  to  H  and  V  and 
whose  base  is  a  circle  on  V;  required  to  assu^ae  a  straight  line  in  a  profile 
plane  and  to  draw  a  plane  tangent  to  the  cone  and  parallel  to  the  line. 

327.  Problem  203.  Given  a  cone  whose  axis  is  oblique  to  H  and  F, 
the  plajie  of  whose  base  is  perpendicular  to  H  but  oblique  to  r,  and 
the  vertical  projection  of  whose  base  is  a  circle  ;  required  to  assume 


Fig.  133 


SURFACES   OF   SINGLE  CUEVATURE 


139 


\ 


a  straight  line  and  to  draw  a  plane  tangent  to  the  cone  and  parallel 
to  the  line. 

328.  Problem  204.  Given  a  cone  whose  axis  is  oblique  to  If  and  V 
and  whose  base  is  a  circle  in  a  plajie  parallel  to  V  and  in  front  of  it; 
required  to  assume  a  straight  line  and  to  draw  a  plaiie  tangent  to 
the  cone  and  parallel  to  the  line. 

329.  Problem  205.  To  draw  a  plane  tangent  to  a  helical  convolute 
at  a  point  on  the  surface. 

Analysis.  Since  the  surface  of  the  helical  convolute  is  of  single 
curvature,  the  rectilinear  element  of  the  surface  through  the  point 
of  tangency  will  be  the  element  of  tangency,  and  therefore  be  a 
line  of  the  required  plane  (see  Section  226). 

Since  the  required  plane  will  be  tangent  to  the  surface  all  along  the 
element  of  tangency,  a  straight  line  tangent  to  the  surface  at  any 
point  of  this  element  of  tangency  will  be  a  line  of  the  required  plane. 

Construction.  Let  the  helical  convolute  be  represented  as  in 
Fig.  134.  A-B  represents  the  axis,  D-E-F-G-K  represents  the 
helical  directrix,  d—l—  ^~<S^- 

m^—n^  ■  •  •   represents  the  f 

base  of  the  surface,  and 
0,  assumed  as  in  Sec- 
tion 250,  represents  the 
given  point  on  the 
surface. 

The  element  R-Q-U 
through  0  is  a  line  of 
the  required  plane  and 
piercesii'at  q,  and 
pierces  V  at  u\  points  in 
the  required  traces. 

Through  q,  draw  S- 
q-Sf  tangent  to  the 
base   of   the    convolute 

at  q,.    This  is  a  line  of  the  plane,  since  it  is  tangent  to  a  curve  of 
the  surface  at  a  point  on  0-Q  (see  Section  226). 

Since  this  tangent  is  also  in  if,  it  is  the  required  horizontal  trace. 
The  straight  line  through  S  and  u'  is  the  required  vertical  trace. 


140 


DESCKIPTIVE  GEOMETRY 


The  element  R-Q-U  is  the  element  of  tangency. 

Check.  Through  0,  or  through  any  point  on  the  element  of  tan- 
gency, draw  a  straight  line  parallel  to  S-s^  and  note  whether  it 
intersects  V  in  the  vertical  trace  already  located. 

330.  Problem  206.  To  draw  a  plane  tangent  to  a  helical  convolute 
and  through  a  point  assumed  without  the  surface. 

Analysis.  A  plane  through  the  given  point  and  intersecting  the 
surface  will  cut  from  the  surface  a  line,  usually  curved,  and  from 
the  required  plane  a  straight  line  running  through  the  given  point 
and  tangent  to  the  curve.  Therefore,  through  the  given  point  pass 
a  plane  cutting  from  the  surface  a  curved  line.    Through  the  given 


\ 


\-"^         /I  h     O,;^/ 


point  and  tangent  to  this  curved  line  draw  a  straight  line.  This 
straight  line  is  a  line  of  the  required  plane  and  will  pierce  H  and  V 
in  points  of  the  required  traces. 

The  element  of  the  surface  through  the  point  of  tangency  just 
determined  on  the  curve  of  the  surface  is  the  element  of  tangency 
and  is  a  line  of  the  required  plane. 

A  straight  line  tangent  to  the  surface  at  any  point  on  the 
element  of  tangency  is  another  line  of  the  required  plane. 

Construction.  Let  the  helical  convolute  be  represented  as  in 
Fig.  135.    A-B  represents  the  axis  and  0  represents  the  given  point. 


SURFACES  OF  SINGLE  CURVATURE  141 

Through  0  pass  the  plane  U  parallel  to  II.  The  plane  U  cuts 
the  helical  directrix  at  the  point  i>,  and  since  the  plane  is  parallel 
to  B,  it  cuts  the  surface  of  the  convolute  in  a  curve  similar  to  the 
curve  cut  from  the  same  surface  by  the  plane  R,  but  starting  from 
the  point  2>. 

The  plane  U  cuts  the  element  whose  horizontal  projection  is 
e-ff  at  a  point  horizontally  projected  at  ^^  where  e^-g^  is  equal  to 
the  rectified  arc  e^-d^  (see  Section  216). 

The  plane  U  cuts  the  element  whose  horizontal  projection  is 
kf-lf  at  a  point  horizontally  projected  at  w^  where  kf-nif  is  equal 
to  the  rectified  arc  k^-e-d^. 

In  the  same  way  other  points  in  which  the  plane  U  cuts  the 
elements  of  the  surface  may  be  found.  The  curve  d^-g-m-n^  •  •  • 
is  then  the  horizontal  projection  of  the  curve  cut  from  the  helical 
convolute  by  the  plane  U. 

Through  o^  draw  o-q^  tangent  to  the  curve  d-g-m^-n^  at  the 
point  q,.  The  line  o^-q,  is  the  horizontal  projection  and  the  line 
o'-q'  is  the  vertical  projection  of  a  straight  line  through  0  and  tan- 
gent to  the  curve  cut  from  the  surface  by  the  plane  U.  0-Q  is 
then  a  line  of  the  required  plane  and  pierces  V  in  r\  a  point  in  the 
required  vertical  trace.  Through  Q  and  tangent  to  the  helical 
directrix  draw  the  element  of  tangency  Q-W-X,  piercing  j^  at  w^ 
and  piercing  V  at  x\  points  in  the  required  traces.  Through  w^ 
draw  the  required  horizontal  trace  S-s^  tangent  to  the  base  of  the 
surface  at  w^.  Through  S  and  x'  draw  the  required  vertical  trace 
S-8\  which  should  also  pass  through  r',  and  check  the  work. 

Check.  Through  any  point  of  the  element  of  tangency  draw  a 
straight  line  parallel  to  S-s^  and  note  whether  it  pierces  V  in  the 
vertical  trace  now  located. 

331.  Problem  207.  To  draw  a  playie  tangent  to  a  helical  convolute 
and  parallel  to  a  given  straight  line. 

Analysis.  Assuming  the  axis  of  the  convolute  perpendicular  to 
H^  with  some  point  on  the  given  line  as  a  vertex  construct  a  cone 
whose  elements  shall  make  with  H  the  constant  angle  which  the 
elements  of  the  convolute  make  with  H. 

Each  element  of  the  convolute  will  then  have  a  corresponding 
element  on  the  cone  to  which  it  will  be  parallel;  and  any  two 


142 


DESCRIPTIVE   GEOMETRY 


planes,  the  one  tangent  to  the  cone  and  the  other  tangent  to  the 
convolute,  along  such  corresponding  parallel  elements,  will  be 
parallel  planes. 

Through  the  given  line  pass  a  plane  tangent  to  this  auxiliary- 
cone.  Such  a  plane  will  be  parallel  to  the  required  plane,  since 
it  contains  the  given  line  and  an  element  on  the  cone  which  is 
parallel  to  the  element  of  tangency  on  the  convolute. 

Construction.  Let  the  helical  convolute  be  represented  as  in 
Fig.  136.  A-B  represents  the  axis  and  Ji-iV^  represents  the  given 
line. 

Through  n'^  the  point  in  which  3I-N  pierces  F,  draw  n'-f  par- 
allel to  e'-d'^  where  e'-d'  is  the  vertical  projection  of  an  element 


Fig.  130 


of  the  convolute  parallel  to  V.  The  elements  of  the  right  circular 
cone  generated  by  the  revolution  of  the  line  n'-f  about  the  verti- 
cal line  through  n'  will  make  the  same  angle  with  H  as  the  elements 
of  the  convolute.  The  circle  whose  center  is  n^  is  the  base  of  the 
cone  and  n'  is  the  vertex  of  the  cone. 

Through  M-N  and  tangent  to  the  cone  draw  the  plane  U  (see 
Section  321).  The  horizontal  trace  U-u^  will  be  tangent  to  the 
circular  base.    The  vertical  trace  U-u'  will  pass  through  w'. 


SURFACES  OF   SINGLE  CURVATURE  143 

Tangent  to  the  base  of  the  convolute  and  parallel  to  U-u^  draw 
S-Sj^  the  required  horizontal  trace.  Through  S  and  parallel  to 
U-u^  draw  S-s'^  the  required  vertical  trace. 

The  element  of  tangency  G-K  may  be  determined  by  drawing 
the  element  of  the  surface  through  the  point  of  tangency  G  on 
the  base. 

Since  another  plane  may  be  drawn  through  M-N  and  tangent  to 
the  cone,  another  plane  may  be  constructed  which  will  answer  the 
conditions  of  the  .problem. 

Since  the  convolute  is  unlimited  in  extent,  having  its  base  in 
the  form  of  a  spiral,  an  unlimited  number  of  planes  answering  the 
conditions  of  the  problem  may  be  constructed. 

Oheok.  Through  any  point,  as  K,  on  the  element  of  tangency 
draw  a  straight  line  K-L  parallel  to  /S-s,,  and  note  whether  it 
pierces  V  in  the  vertical  trace  already  located. 


CHAPTER  XIY 

DETERMINATION  OF  PLANES  TANGENT  TO   SURFACES 
OF  DOUBLE  CURVATURE 

332.  General  Instructions.  By  Section  226  straight  lines  which 
are  tangent  to  a  surface  of  double  curvature  at  a  point  on  the  sur- 
face, lie  in  a  plane  tangent  to  the  surface  at  this  point.  There- 
fore through  the  point  of  tangency  draw  two  planes  cutting  from 
the  surface  two  simple  curved  lines  intersecting  at  the  point 
of  tangency.  Tangent  to  these  curved  lines  at  the  point  of  tan- 
gency draw  two  straight  lines  which  shall  determine  the  required 
plane. 

When  the  given  surface  is  one  of  revolution,  it  will  be  found 
convenient  to  use  as  the  cutting  planes  the  meridian  plane  and  a 
plane  perpendicular  to  the  axis. 

Sometimes  an  auxiliary  surface  may  be  passed  tangent  to  the 
given  surface  at  the  point  of  tangency;  then  of  course  a  plane  tan- 
gent to  the  auxiliary  surface  at  the  point  of  tangency  will  also  be 
tangent  to  the  given  surface  at  the  point  of  tangency. 

333.  Problem  208.  To  draw  a  plane  tangent  to  a  sphere  at  a  point 
on  the  surface. 

Analysis  1.    See  Section  332. 

Analysis  2.  Draw  a  plane  perpendicular  to  the  radius  of  the 
sphere  at  the  point  of  tangency. 

Construction.  See  Fig.  137.  Let  C  represent  the  sphere  and  let 
0,  assumed  as  in  Section  287,  represent  the  point  on  the  surface. 

By  Analysis  1  draw  through  0  a  vertical  meridian  plane  T,  cut- 
ting the  sphere  in  a  great  circle.  Revolve  T  about  its  horizontal 
trace  into  H.  The  center  of  the  great  circle  will  fall  at  Cjj  and  the 
point  of  tangency  will  fall  at  Ojj.  Through  Ojj  draw  Ojj-a^  tangent 
to  the  circle  Cjj.  This  last  line  is  the  revolved  position  of  a  tangent 
to  the  sphere  at  the  point  0.  This  tangent  in  true  position  pierces 
H  at  a,,  a  point  in  the  required  horizontal  trace. 

144 


SURFACES  OF  DOUBLE  CURVATURE 


145 


Through  0  pass  another  plane  TJ  parallel  to  //.  This  plane  cuts 
the  sphere  in  a  small  circle  whose  horizontal  projection  is  o,-o„-5, 
and  whose  vertical  projection  is  o^^-V .  Through  0  draw  0-D  tan- 
gent to  the  small  circle  at  0.  0-D  is  another  line  of  the  required 
plane,  and  pierces  V  at  d! ^ 
a  point  in  the  required 
vertical  trace. 

Since  0-D  is  a  line  of 
the  required  plane  and  par- 
allel to  H^  its  horizontal 
projection  o^-d^  must  be 
parallel  to  the  required  hor- 
izontal trace  (see  Section 
42).  Therefore  through  a^ 
and  parallel  to  o^-d^  draw 
the  required  horizontal 
trace  iS-s, .  Through  S  and 
d^  draw  the  required  ver- 
tical trace  S-8\ 

Check.  The  traces  of  the  required  plane  should  be  perpendicular 
to  the  projections  of  the  radius  C-0. 

334.  Problem  209.    Solve  the  above  problem  hy  Analysis  2. 

335.  Problem  210.  G-iven  a  sphere  with  center  in  G-L;  required 
to  assume  a  point  upon  the  surface  and  to  pass  a  plane  tangent  to  the 
surface  at  this  point, 

336.  Problem  211.  Given  a  sphere  in  the  second  quadrant  with  its 
center  equidistant  from  H  and  V ;  required  to  assume  a  point  upon 
the  surface  and  to  pass  a  plane  tangent  to  the  surface  at  this  point. 

337.  Problem  212.  To  draw  a  plane  tangent  to  an  ellipsoid  of 
revolution  at  a  point  on  the  surface. 

Analysis  1.    See  Section  332. 

Ajialysis  2.  Determine  the  meridian  curve  of  the  surface  which 
contains  the  given  point,  and  draw  a  rectilinear  tangent  to  the 
curve  at  this  point.  If  this  tangent  is  revolved  about  the  axis  of 
the  ellipsoid  as  an  axis,  it  will  generate  a  conical  surface  which 
will  be  tangent  to  the  surface  of  the  ellipsoid  in  the  circumference 
of  a  circle  containing  the  point  of  tangency  (see  Section  284). 


146 


DESCKIPTIVE   GEOMETRY 


A  plane  tangent  to  this  cone  along  the  element  passing  through 

the  given  point  of  tangency  will  also  be  tangent  to  the  ellipsoid 

at  the  same  point. 

Construction.    Let  the  ellipsoid  be  represented  as  in  Fig.  138, 

where  A~B  represents  the  long  axis  and  where  0  represents  the 

point  assumed  on  the  surface. 

Following  Analysis  1,  pass  a  meridian  plane  T,  containing  the 

point  0  and  cutting  the  surface  in  a  meridian  ellipse.    Revolve  T 

about  A-B  as  an  axis  until  it  is  parallel  to  V.    The  point  0  will  then 

be  vertically  projected 
at  o"  and  the  meridian 
ellipse  will  be  vertically 
projected  in  the  vertical 
projection  of  the  ellipsoid. 
Tangent  to  the  ellipse 
a'-o"-b'-f  at  the  point  o" 
draw  d'-o"-e".  This  line 
is  the  vertical  projection 
of  the  revolved  position 
of  a  straight  line  D-E^ 
which  in  true  position  is 
tangent  to  the  surface  at 
0,  and  therefore  in  the 
required  plane.  In  re- 
volved  position    D-E 

pierces  H  at  e,, ,  but  in  true  position,  since  D  is  in  the  axis,  it  will 

pierce  H  at  e^. 

Through  0  pass  another  auxiliary  plane  U  perpendicular  to  the 

axis  A-B.    This  plane  cuts  the  ellipsoid  in  a  circle  whose  horizontal 


—B- 


Fig.  138 


projection  is/,- 


-o,—o. 


and  whose  vertical  projection  is/'- 


The  remainder  of  the  construction  corresponds  with  that  given 
in  connection  with  the  sphere  (see  Section  333). 
Check.    Construct  a*ccording  to  Analysis  2. 

338.  Problem  213.    Solve  the  above  problem  by  Analysis  2. 

339.  Problem  214.  Given  an  ellipsoid  of  revolution  with  the  long 
axis  perpendicular  to  V;  required  to  assume  a  point  upon  the  surface 
and  to  draw  a  plane  tangent  to  the  surface  at  this  point. 


SURFACES  OF  DOUBLE  CURVATURE  147 

340.  Problem  215.  Given  an  ellipsoid  of  revolution  with  the  long 
axis  parallel  to  G—L;  required  to  assume  a  point  upon  the  surface 
and  to  draw  a  plane  tangent  to  the  surface  at  this  point. 

341.  Problem  216.  To  draw  a  plane  tangent  to  any  surface  of 
revolution. 

Analysis.  The  directions  given  in  connection  with  the  solution 
of  the  foregoing  problems  will  be  sufficient  for  the  solution  of  all 
problems  relative  to  tangency  of  planes  to  surfaces  of  revolution. 

342.  Problem  217.  To  draw  a  plane  containing  a  given  straight 
line  and  tangent  to  a  given  sphere. 

Analysis  1.  After  the  required  plane  is  constructed,  an  auxiliary 
plane  through  the  center  of  the  sphere  and  perpendicular  to  the 
given  line  will  cut  from  the  sphere  a  great  circle,  from  the  line  a 
point,  and  from  the  required  plane  a  straight  line  passing  through 
the  point  and  tangent  to  the  great  circle.  Therefore,  through  the 
center  of  the  sphere  and  perpendicular  to  the  given  line,  pass  a 
plane  cutting  the  sphere  in  a  great  circle  and  cutting  the  line  in 
a  point.  Through  this  point  and  tangent  to  the  circle  draw  a 
straight  line.  This  line  together  with  the  given  line  will  determine 
the  required  plane. 

Analysis  2.  If  through  any  point  of  the  given  line  a  series  of 
rectilinear  tangents  to  the  sphere  be  drawn,  they  will  form  the 
elements  of  a  cone  whose  axis  will  contain  the  point  on  the  line 
and  the  center  of  the  sphere.  The  surface  of  this  cone  will  be 
tangent  to  the  sphere  in  the  circumference  of  a  small  circle  whose 
plane  will  be  perpendicular  to  the  axis  of  the  cone. 

A  plane  tangent  to  the  cone  must  also  be  tangent  to  the  sphere. 
Therefore  a  plane  through  the  given  line  and  tangent  to  the  cone 
will  be  the  required  plane. 

Analysis  3.  If  from  each  of  two  points  on  the  given  line  a  series 
of  rectilinear  tangents  to  the  sphere  be  drawn,  they  will  form  the 
elements  of  two  cones  whose  axes  will  contain  the  points  on  the 
line  and  the  center  of  the  sphere,  and  whose  surfaces  will  be  tangent 
to  the  sphere  in  the  circumferences  of  small  circles,  which  may  be 
taken  as  the  bases  of  the  cones.  The  planes  of  these  small  circles 
will  be  perpendicular  to  the  axes  of  their  respective  cones,  and  will 
intersect  in  a  straight  line  which  will  be  a  chord  of  the  sphere. 


148 


DESCKIPTIVE  GEOMETEY 


The  circumferences  of  these  two  circles  will,  as  a  rule,  intersect 
in  two  points  on  the  surface  of  the  sphere,  which  will  be  the 
points  in  which  the  chord  of  the  sphere  intersects  the  surface  of 
the  sphere,  and  which  will  be  the  points  at  which  the  required 
planes  will  be  tangent  to  the  sphere.  For  a  straight  line  drawn 
through  the  vertex  of  either  cone  and  one  of  these  points  of  tan- 
gency  is  an  element  of  this  cone  and  will  therefore  be  tangent  to 
the  sphere  at  this  point  of  tangency.  A  straight  line  through  the 
vertex  of  the  other  cone  and  this  same  point  of  tangency  is  an 


Fig.  139 

element  of  the  other  cone  and  will  therefore  be  tangent  to  the 
sphere  at  this  same  point  of  tangency. 

The  plane  of  these  two  tangents  will  contain  the  given  line  and 
will  be  tangent  to  the  sphere  at  this  point  of  tangency. 

Construction  1.  In  Fig.  139  let  C  represent  the  center  of  the 
sphere  and  let  M-N  represent  the  given  line. 

Produce  M-N  to  intersect  //  at  a,  and  to  intersect  V  at  h\  points 
in  the  required  traces. 

Following  Analysis  1,  pass  the  plane  U  through  the  point  C  and 
perpendicular  to  M-N  (see  Section  165).  Find  the  point  D  in 
which  U  is  intersected  by  M-N  (see  Section  151). 

Revolve  the  plane  U  about  its  horizontal  trace  into  H.  D  will 
fall  at  djj  and  the  center  of  the  great  circle  cut  from  the  sphere  by 


SURFACES   OF  DOUBLE  CURVATURE 


149 


U  will  fall  at  Cj^.  Through  d^  draw  ^jr^n-ft  tangent  to  the  circle 
Cjj  at  Cjf.  The  straight  line  of  which  ^ji-d^f-fj  is  the  revolved  posi- 
tion passes  through  D  and  is  tangent  to  the  great  circle  cut  from 
the  sphere  by  ?7,  and  is  therefore  a  line  of  the  required  plane. 
Produce  this  line  to  meet  H  in  /,.  Through  /,  and  a^  draw  S-s^^ 
the  required  horizontal  trace.  Through  S  and  h'  draw  S-s',  the 
required  vertical  trace. 

Chech,  Assume  any  point,  as  (7,  in  the  vertical  trace  of  U,  After 
the  revolution  of  U^  G  will  take  the  position  ^^^,  and  the  vertical 
trace  U-u'  will  take  the  position  U-gji,  The  point  kjj^  the  point  in 
which  U-gjj  intersects  e^-/,,  is  the  revolved  position  of  the  point 
in  which  E-F  crosses  the  vertical  trace  of  Z7,  or,  in  other  words, 


Fig.  140 


kji  is  the  revolved  position  of  the  point  in  which  E-F  pierces  F,  a 
point  necessarily  in  the  vertical  trace  of  the  required  plane. 

The  true  position  of  this  point  K  must  be  upon  U-u'  at  a  dis- 
tance from  U  equal  to  U-k^^^  and  should  be  on  S-s'  already  located. 

Since  from  djj  another  rectilinear  tangent  to  the  circle  c,i  may  be 
drawn,  another  plane  answering  the  conditions  of  the  problem  may 
be  constructed. 

Construction  2.  In  Fig.  140  let  C  represent  a  sphere  whose  cen- 
ter is  in  G-L  and  let  M-N  represent  the  given  line. 


150  DESCRIPTIVE  GEOMETRY 

Produce  M-N  to  intersect  H  in  a,  and  to  intersect  V  in  h\  points 
in  the  required  traces. 

Following  Analysis  2,  take  the  point  ^  —  the  point  in  which  M-N 
pierces  H — as  the  vertex  of  a  cone  whose  elements  are  tangent  to 
the  sphere.  The  lines  a-d^  and  a-e^^  each  tangent  to  the  hori- 
zontal projection  of  the  sphere,  will  represent  the  horizontal 
projections  of  the  extreme  elements  of  the  cone.  The  line  a-c^ 
represents  the  horizontal  projection  of  the  axis  of  the  cone,  and 
U-d-e-iij  represents  the  horizontal  trace  of  the  plane  of  the  circle 
of  tangency,  which  circle  may  be  taken  as  the  base  of  the  cone. 

Determine  the  point  F  in  which  M-N  intersects  U.  Revolve  U 
about  its  horizontal  trace  U-u^  as  an  axis  into  H.  F  will  fall  at 
fjj^  and  the  circle  of  tangency  will  fall  at  d-hjj-e^.  Through  fjj 
draw  fii-kji-lj  tangent  to  the  circle  d^-kj^-e,  at  the  point  kj^. 

This  line  is,  by  Section  321,  the  revolved  position  of  a  line  of 
the  plane  containing  the  line  M-N  and  tangent  to  the  cone,  and 
is  therefore  a  line  of  the  required  plane. 

In  true  position  this  line  F-K-L  pierces  H  at  Z,.  Through  a, 
and  If  draw  the  required  horizontal  trace  S-s^.  Through  >S^  and  h' 
draw  the  required  vertical  trace  S-s'. 

Chech.  The  required  plane  is  tangent  to  the  cone  along  the 
element  A-K.,  and  since  this  element  is  tangent  to  the  sphere  at 
the  point  A',  K  is  the  point  at  which  the  required  plane  is  tangent 
to  the  sphere.  The  required  plane  must  then  be  perpendicular  to 
the  radius  K-C.  The  projections  of  K  in  true  position  are  h^  and 
lc\  where  the  distance  of  k'  from  G-L  is  equal  to  k-kj^.  Therefore 
S-Sf  and  S-s'  should  be  perpendicular  respectively  to  k^-c,  and  k'-c'. 

Since  from  the  point  fjj  another  tangent  to  the  circle  d-k^j-e^ 
may  be  drawn,  another  plane  answering  the  conditions  of  the 
problem  may  be  constructed. 

Construction  3.  See  Fig.  141.  C  represents  the  center  of  the 
sphere  and  M-N  represents  the  given  line. 

M-N  pierces  H  at  a^  and  pierces  V  at  5',  points  in  the  required 
traces. 

Following  Analysis  3,  take  for  the  vertices  of  the  cones  the 
points  D  and  E^  which  are  assumed  upon  M-N  in  such  a  way  that 
i>  is  at  the  same  distance  below  H  as  the  center  of  the  sphere. 


SUEFACES  OF  DOUBLE  CURVATUKE 


151 


and  that  E  is  at  the  same  distance  back  of    V  as  the  center  of 
the  sphere. 

The  axis  of  the  cone  D  is  D-C  and  by  construction  is  parallel 
to  H.  The  axis  of  the  cone  E  is  E-C  and  by  construction  is 
parallel  to  V. 

To  represent  the  horizontal  projection  of  the  cone  D,  draw 
through  (ip  and  tangent  to  the  horizontal  projection  of  the  sphere, 
the  two  lines  d-fj  and 
d,-g,. 

To  represent  the  verti- 
cal projection  of  the  cone 
E^  draw  through  e',  and 
tangent  to  the  vertical  pro- 
jection of  the  sphere,  the 
two  lines  e'-k^  and  e^-V. 

The  base  of  the  cone  D 
is  a  circle  whose  plane 
is  perpendicular  to  H  and 
whose  horizontal  projec- 
tion is  f-g,.  The  base  of 
the  cone  ^  is  a  circle 
whose  plane  is  perpendic- 
ular to  V  and  whose  ver- 
tical projection  is  k'-V. 

The  planes  of  these 
bases  will  intersect  in  a 
straight  line  whose  hori- 
zontal projection  will  fall  on  /,-^,  and  whose  vertical  projection 
will  fall  on  k'-V, 

This  line  of  intersection  will  pierce  the  surface  of  the  sphere  in 
the  two  points  in  which  the  circumferences  of  the  two  bases  inter- 
sect, and  which  by  analysis  are  the  points  of  tangency  sought. 

To  find  these  points  of  tangency  revolve  the  plane  of  the  base 
of  the  cone  whose  vertex  is  D  about  its  horizontal  trace /,-^,  into 
H.  The  circular  base  whose  center  is  0  will  take  the  position  of 
the  circle  whose  center  is  Ojj,  where  o-Ojj  is  equal  to  the  distance 
of  (?'  from  G-L.    The  intersection  of  the  planes  of  the  two  bases 


Fig.  141 


152  DESCRIPTIVE  GEOMETRY 

will  fall  at  qif-Pfi  where  p,  represents  the  point  in  which  this  line 
of  intersection  in  true  position  pierces  //  and  where  Q  represents 
any  point  assumed  on  the  line. 

The  points  Vjj  and  Ujj  are  then  the  revolved  positions  of  the  two 
points  of  tangency  sought. 

Through  Vj,  and  tangent  to  the  circle  Ojj  draw  rjj-w^.  This  is 
the  revolved  position  of  a  line  of  the  required  plane,  since  it  is  the 
revolved  position  of  a  straight  line  tangent  to  the  sphere  at  the 
point  at  which  the  required  plane  is  to  be  tangent.  The  line 
R-W  in  true  position  pierces  //  at  w^  and  pierces  V  at  x\  where 
x-x'  is  equal  to  Xj-^Xjj  measured  on  a  straight  line,  x-x^^  perpen- 
dicular to  Wf-Xf. 

Through  a^  and  w^  draw  the  required  horizontal  trace  S-s^, 
Through  b'  and  x'  draw  the  required  vertical  trace  S-s'. 

Check.  The  two  traces  should  cross  G-L  at  the  same  point.  Or, 
the  two  traces,  S-s^  and  S-s\  should  be  perpendicular  respectively 
to  the  two  projections  of  the  radius  R-C. 

Since  the  line  Q-F  intersects  the  surface  of  the  sphere  in  another 
point  U,  another  plane  answering  the  conditions  of  the  problem 
may  be  constructed. 

343.  Problem  218.  Given  a  sphere  whose  center  is  in  G-L^  and 
given  a  straight  line  parallel  to  G-L  and  in  the  third  quadrant; 
required  to  draw  a  plane  contaiyiing  the  line  and  tangent  to  the  sphere, 

344.  Problem  219.  Given  a  sphere  whose  center  is  in  G-L^  and 
given  a  straight  line  situated  in  a  profile  plane  and  in  the  third 
quadrant;  required  to  pass  a  plane  through  the  line  and  tangent  to 
the  sphere. 

345.  Problem  220.  Given  a  sphere  in  the  second  quadrant  equi- 
distant from  H  and  F,  also  given  a  straight  line  in  the  first  quadrant; 
required  to  pass  a  plane  through  the  line  and  tangent  to  the  sphere. 


CHAPTER  XV 

INTERSECTION  OF  SURFACES  BY  LINES 

346.  General  Instructions.  If  the  given  line  is  a  straight  line  or 
a  curved  line  of  single  curvature,  pass  an  auxiliary  plane  through 
the  line  and  determine  its  intersection  with  the  given  surface. 
The  point  or  points  in  which  the  given  line  intersects  the  lines 
cut  from  the  surface  by  the  auxiliary  plane  will  be  the  points 
sought. 

Since  an  infinite  number  of  planes  may  be  drawn  through  a 
straight  line,  it  will  be  wise,  in  connection  with  straight  lines,  to 
use  those  auxiliary  planes  which  will  cut  from  the  surface  the 
simplest  lines.  If  the  given  line  is  a  curved  line  of  single  curva- 
ture the  plane  of  the  curve  must  be  used  as  the  auxiliary  plane. 

347.  On  the  Character  of  Lines  in  which  Planes  intersect  Surfaces. 
Since  prisms  and  pyramids  have  plane  surface  faces,  the  intersec- 
tions of  these  faces  by  planes  will  always  be  straight  lines  what- 
ever the  position  of  the  cutting  plane. 

Since  the  elements  of  cylindrical  surfaces  are  parallel  straight 
lines,  the  intersections  of  such  surfaces  by  planes  which  are  par- 
allel to  the  elements  are  parallel  straight  lines. 

Since  the  elements  of  conical  surfaces  all  pass  through  the  ver- 
tex, the  intersections  of  such  surfaces  by  planes  which  contain  the 
vertex  will  be  straight  lines  passing  through  the  vertex. 

The  intersection  of  a  sphere  by  a  plane  is  always  a  circle.  If 
the  cutting  plane  contains  the  center  of  the  sphere,  the  circle  is  a 
great  circle. 

348.  Problem  221.  To  find  the  point  in  which  a  given  straight  line 
intersects  a  given  plane  {see  Section  151). 

349.  Problem  222.  To  find  the  points  in  which  a  given  straight 
line  intersects  the  surface  of  a  square  prism. 

Analysis.  Pass  the  auxiliary  plane  through  the  line  and  parallel 
to  the  edges  of  the  prism. 

153 


154 


DESCRIPTIVE  GEOMETRY 


Fig.  142 


Construction.  Let  the  prism  be  assumed  with  its  edges  perpen- 
dicular to  //,  as  shown  in  Fig.  142,  and  let  M-N  represent  the 
given  line. 

The  auxiliary  plane  is  taken  perpendicular  to  H  and  intersects 
the  surface  in  two  lines,  A-B  and  D-B,  each  parallel  to  the  edges. 

These  lines  are  crossed  by  3f-N  at  X 
and  r,  the  two  points  required. 

350.  Problem  223.  Find  the  points  in 
which  a  straight  line  intersects  the  sur- 
face of  a  prism  whose  edges  are  oblique 
to  II  and  V. 

351.  Problem  224.  Find  the  points  in 
which  a  straight  line  intersects  the  sur- 
face of  an  hexagonal  prism. 

352.  Problem  225.  Given  a  hollow 
square  prism  ;  required  to  find  the  points 
in  which  a  given  straight  line  intersects 

the  outer  and  the  inner  surface  of  the  prism. 

353.  Problem  226.    To  find  the  points  iyi  which  a  given  straight  line 
intersects  the  surface  of  a  square 
pyramid. 

Analysis.  Pass  the  auxiliary 
plane  through  the  line  and  the 
vertex  of  the  pyramid. 

Construction.  Let  the  pyramid 
be  represented  as  in  Fig.  143, 
and  let  M-N  represent  the  given 
line.  Produce  M-JV  to  meet  the 
plane  of  the  base  of  the  pyramid 
at  B.  Through  the  vertex  A  and 
any  point  D  of  the  line  M-K 
draw  a  straight  line  and  produce 
it  to  meet  the  plane  of  the  base 

at  F.    The  straight  line  through  B  and  F  is  the  intersection  of 
the  auxiliary  plane  and  the  plane  of  the  base  of  the  pyramid. 

This  auxiliary  plane  intersects  the  pyramid  in  two  lines,  A-F  and 
A-G.    J/-iV^  crosses  these  lines  at  A"  and  F,  the  points  required. 


Fig.  143 


INTEKSECTION  OF  SURFACES   BY  LIKES 


155 


354.  Problem  227.  Find  the  points  in  which  a  given  straight  line 
intersects  the  surface  of  an  oblique  hexagonal  pyramid. 

355.  Problem  228.  Find  the  points  in  which  a  given  straight  line 
intersects  the  surface  of  the  frustum  of  a  square  pyramid. 

356.  Problem  229.  Given  a  hollow  square  pyramid ;  required  to 
find  the  points  in  which  a  given  straight  line  intersects  the  outer  and 
the  inner  surface  of  the  pyramid. 

357.  Problem  230.  To  find  the  points  in  which  a  given  straight 
line  intersects  the  surface  of  a  cylinder. 

Analysis.  Pass  the  auxiliary  plane  through  the  line  and  parallel 
to  the  elements  of  the  cylinder,  or,  what  is  the  same  thing,  parallel 
to  the  axis  of  the  cylinder. 

Construction.  Let  the  cylinder 
be  represented  as  in  Fig.  144, 
and  let  M-N  represent  the  given 
line.  Produce  M-N  to  pierce  H 
at  dj.  Through  iV,  any  point  on 
M-N^  draw  N-E  parallel  to  the 
axis  of  the  cylinder  and  produce 
it  to  pierce  IT  at  e,.  The  straight 
line  U-df-Cf-u^^  determined  by 
the  points  d^  and  e,,  is  the  hcfri- 
zontal  trace  of  the  auxiliary 
plane. 

This  plane  U  cuts  the  cylinder  in  two  elements,  F-X  and 
G-Y,  which  are  intersected  by  M-N  at  X  and  F,  the  two  points 
required. 

358.  Problem  231.  Given  a  right  circular  cylinder  in  the  third 
quadrant  with  axis  parallel  to  G-L;  required  to  find  the  points  in 
which  a  given  straight  line  intersects  the  surface  of  the  cylinder. 

359.  Problem  232.  Given  a  right  circular  cylinder  in  the  third 
quadrant  with  axis  perpendicular  to  H ;  required  to  find  the 
points  in  which  a  given  straight  line  intersects  the  surface  of  the 
cylinder. 

360.  Problem  233.  Given  a  hollow  cylinder ;  required  to  find  the 
points  in  which  a  given  straight  line  intersects  the  inner  and  the 
outer  surface  of  the  cylinder. 


Fig.  144 


156 


DESCEIPTIVE  GEOMETKY 


Fig.  145 


361.  Problem  234.  To  find  the  points  m  which  a  given  straight 
line  intersects  the  surface  of  a  cone. 

Analysis.    Pass  the  auxiliary  plane   through  the  line  and  the 

vertex  of  the  cone. 

\|i  Construction.    Let   the    cone    be 

l\      represented  as  in  Fig.  145,  and  let 

'^  M-N  represent  the  given  line. 

Produce  M-N  to  pierce  H  at  5,. 
Through  J,  the  vertex  of  the  cone, 
and  parallel  to  the  given  line,  draw 
A-D  piercing  H  at  d^ .  U-d-h^-Uf 
is  the  horizontal  trace  of  the  aux- 
iliary plane.  This  plane  intersects 
the  surface  of  the  cone  in  two  ele- 
ments, A-E  and  A-F,  which  are 
cut  by  M-N  in  X  and  F,  the  two 
points  required. 

362.  Problem  235.  Find  the  j^oints  in  which  a  giveyi  straight 
line  intersects  the  surface  of  an  inverted  cone  situated  in  the  third 
quadrant. 

363.  Problem  236.    Find  the  jyoints  in  which  a  given  straight  line 
intersects  the  surface  of  the  frustun\ 
of  a  cone. 

364.  Problem  237.  Given  a  hollow 
cone;  required  to  find  the  points  in 
which  a  given  straight  line  intersects 
the  inner  and  the  outer  surface  of 
the  cone. 

365.  Problem  238.  To  find  the 
points  in  which  a  given  straight  liiie 
intersects  the  surface  of  a  sphere. 

Analysis.  Since  the  intersection 
of  a  sphere  by  any  plane  is  a  circle, 
the  auxiliary  plane  may  be  passed 
through  the  line  at  random.  ^ig.  140 

Construction.  Let  the  sphere  be  represented  as  in  Fig.  146,  and 
let  M-N  represent  the  given  line. 


INTERSECTION  OF  SURFACES  BY  LINES  157 

Take  as  the  auxiliary  plane  the  horizontal  projecting  plane  of 
the  given  line.  This  plane  U  cuts  the  sphere  in  a  small  circle 
whose  center  is  A  and  whose  diameter  is  B-D. 

To  determine  the  points  in  which  M-N  intersects  this  circle, 
revolve  the  plane  U  about  its  horizontal  trace  into  H,  The  center 
A  of  the  circle  cut  from  the  sphere  falls  at  a^,  and  the  line  M-N 
falls  at  mj^-iij^.  The  points  Xjj  and  ^^,  the  points  in  which  mjj-rijj 
intersects  the  circumference  of  the  circle  a^,  are  the  required  points 
in  revolved  position.  The  points  X  and  F,  whose  projections  are 
{x,^  x')  and  (y^  y')  respectively,  are  the  required  points  in  true 
position. 

The  problem  may  be  as  easily  solved  by  using  the  vertical  pro- 
jecting plane  of  M-N  as  the  auxiliary  plane. 

366.  Problem  239.  Given  a  hollow  hemisphere ;  required  to  find 
the  points  in  which  a  given  straight  line  iritersects  the  inner  and  the 
outer  siCrface. 

Analysis.  Take  as  the  auxiliary  plane  one  of  the  projecting 
planes  of  the  given  line,  as  in  the  preceding  problem.  Find  the 
two  semicircles  in  which  this  plane  intersects  the  outer  and  inner 
surfaces  of  the  hemisphere.  The  points  in  which  the  given  line 
intersects  these  semicircles  will  be  the  required  points. 


CHAPTER  XYI 

INTERSECTION  OF  SURFACES  BY  PLANES 

367.  General  Instructions.  Pass  a  series  of  auxiliary  planes 
intersecting  both  the  given  plane  and  the  given  surface  in  lines. 
The  points  in  which  these  two  sets  of  lines  intersect  will  be 
common  to  both  the  plane  and  the  surface,  and  will  therefore 
be  points  in  their  line  of  intersection. 

Pass  the  auxiliary  planes  in  such  a  way  as  not  only  to  cut  from 
the  given  surface  the  simplest  lines,  but  also  in  such  a  way  as  to 
simplify  the  work  of  construction. 

If  the  given  surface  is  that  of  a  prism  or  of  a  pyramid,  it  will  be 
found  convenient  to  pass  the  auxiliary  planes  through  the  edges. 

If  the  given  surface  is  one  of  revolution,  it  will  be  found  con- 
venient to  pass  the  auxiliary  planes  perpendicular  to  the  axis  of 
revolution,  since  such  planes  will  cut  circumferences  of  circles  from 
the  surface. 

368.  Problem  240.  To  find  the  line  in  which  one  plane  intersects 
another  {see  Section  14-6), 

369.  Problem  241.  To  find  the  intersection  of  a  triarigular  prism 
hy  a  plane,  to  find  the  true  size  of  the  intersection,  and  to  develop 
the  surface  of  the  prism. 

Case  1.    To  find  the  intersection. 

Analysis.  Pass  the  auxiliary  planes  through  the  edges  of  the 
prism. 

Construction.  In  Fig.  147  let  A-B-D — E-F-G  represent  a  prism 
with  edges  perpendicular  to  H,  and  let  S  represent  the  cutting 
plane. 

The  plane  of  the  face  A-D-G-E,  which  is  the  plane  U,  contains 
the  two  edges  A-E  and  D-G.  This  plane,  which  may  be  used  as 
an  auxiliary  plane,  cuts  S  in  the  line  K-L,  crossing  A-E  and  D-G 
at  X  and  Z  respectively.  X-Z  is  the  line  of  intersection  between 
the  plane  S  and  the  face  A-D-G-E. 

158 


INTERSECTION   OF   SURFACES  BY  PLANES 


159 


Pass  another  auxiliary  plane,  W\  through  the  edge  B-F,  and 
locate  the  point  Y  in  which  the  edge  B-F  is  cut  by  S. 

S  intersects  the  face  D-B-F-G  in  the  line  Y-Z^  and  intersects 
the  face  A-B-F-E  in  the  line  X-Y.  The  required  intersection  is 
X-Y-Z. 

In  this  case  the  auxiliary  planes  are  perpendicular  to  H^  since 
the  edges  of  the  prism  are  perpendicular  to  H,  The  method  of 
construction  is  the  same  when  the  edges  are  oblique  to  i/,  but 
the  auxiliary  planes  will  then  be  oblique  to  H. 

Case  2.  To  find  the  true  size  of 
the  intersectio7i. 

Analysis  and  Construction.  Re- 
volve the  plane  >S,  in  Fig.  147, 
about  S-8^  as  an  axis  into  H.  The 
three  vertices  X,  F,  and  Z  will  fall 
respectively  at  a:^,  yj^^  and  Zj^. 
The  true  size  of  the  intersection  is 
Xjj-yj^-Zjj  (see  Section  87). 


Fig.  147 


Fig.  148 


The  triangle  X-Y-Z  is  the  exact  pattern  of  the  opening  which 
would  be  necessarily  made  in  the  plane  S  in  order  to  allow  the 
prism  to  pass  through. 

Case  3.    To  develop  the  surface  (see  Section  228). 

Analysis.  Take  the  plane  of  the  face  A-B-F-E  as  the  plane  of 
development. 

Starting  with  this  face  in  the  plane,  roll  the  prism  from  left  to 
right,  bringing  first  the  face  B-D-G-F  and  later  the  face  A-D-G-E 
into  coincidence  with  the  original  plane. 

Construction.  Fig.  148  shows  how  these  faces  will  appear  after 
development. 


160  DESCEIPTIVE  GEOMETRY 

Since  the  plane  of  the  base  of  the  prism  is  perpendicular  to  the 
edges  of  the  prism,  the  straight  line  A-B-D-A\  where  A-B,  B-D, 
and  D-A'  are  made  equal  respectively  to  a-b^^  h-d^,  and  d^-a,  of 
Fig.  147,  may  represent  the  development  of  the  base  of  the  surface. 

The  edges  of  the  prism  will,  in  development,  take  the  positions 
A-E,  B-F^  D-G^  and  A'-E\  all  perpendicular  to  A-A'  and  all 
equal  to  the  altitude  of  the  prism. 

The  rectangle  A-A'-E'-E  (Fig.  148)  represents  the  developed 
prism. 

The  vertices  of  the  triangle  of  intersection,  X-  Y-Z  of  Fig.  147, 
will  in  development  take  the  positions  A',  F,  Z,  and  X'  (Fig.  148) 
where  A-X^  B-  F,  D-Z^  and  A'-X'  are  made  equal  respectively  to 
a^-x',  h'-y',  d'-z\  and  a'-x'  of  Fig.  147.  The  line  X-Y-Z-X^  of 
Fig.  148  represents  the  line  of  intersection  in  development. 

The  plane  surface  E-X-Y-Z-X'-E'-E  (Fig.  148)  is  the  temp- 
let or  pattern  of  that  portion  of  the  surface  of  the  prism  below 
the  plane  S^  and  might  be  used  in  the  construction  of  a  duplicate 
prism  to  take  the  place  of  that  portion  of  the  prism  now  there. 

370.  Problem  242.  Given  a  triangular  prism  whose  edges  are 
oblique  to  IT  and  F",  and  given  a  cutting  plane  perpendicular  to  the 
edges  of  the  prisyn  ;  required  to  find  the  intersection^  the  true  size  of 
this  intersectio7i^  and  the  development  of  the  surface  of  the  prism. 

Note.    In  developing  this  surface  use  the  cutting  plane  as  a  basal  plane. 

371.  Problem  243.  Given  a  square  prism  whose  edges  are  parallel 
to  G—Ly  and  given  a  cutting  plane  oblique  to  H  and  V ;  required  to 
find  the  intersection^  the  true  size  of  this  intersection,  and  the  devel- 
opment of  the  surface  of  the  prism. 

Note.  In  developing  this  surface  use  an  auxiliary  profile  plane  as  a  basal 
plane. 

372.  Problem  244.  Given  an  hexagonal  prism  whose  edges  are 
oblique  to  H  and  F,  and  given  a  cutting  plane  parallel  to  H  ;  required 
to  find  the  intersection.,  the  true  size  of  this  intersection.,  and  the  devel- 
opment of  the  surface  of  the  prism. 

Note.  In  developing  this  surface  use  an  auxiliary  cutting  plane  perpen- 
dicular to  the  edges  of  the  prism,  as  a  basal  plane. 


INTERSECTION  OF  SURFACES  BY  PLANES 


161 


373.  Problem  245.  To  find  the  iyitersection  of  a  square  pyramid  hy 
a  plane,  to  find  the  true  size  of  the  intersection,  and  to  develop  the 
surface  of  the  pyramid. 

Case  1 .    To  find  the  intersection. 

Analysis.  Pass  the  auxiliary  planes  through  the  edges  of  the 
pyramid. 

Construction.  In  Fig.  149  let  A-B-D-E-F  represent  the  pyra- 
mid and  let  S  represent  the  cutting  plane. 

Pass  an  auxiliary  plane  T  through  the  two  edges  A-B  and  A-E. 
This  plane  intersects  S  in  G-K,  which  crosses  the  edges  A-B  and 

A-E  at  X  and  Z  respectively. 

Pass  another  auxiliary  plane  U 

through  the  two  edges  A-D  and 

A-F.  U  intersects  S  in  L-  0,  which 

crosses  the  two  edges  A-D  and  A-F 

at  Fand  W  respectively.    X-Y- 

W  is  the  required  intersection. 

In    this    particular 

•L     case,   on    account    of 

the  character  and 


Fig.  149 


Fig.  150 


position  of  the  pyramid,  the  auxiliary  planes  T  and  U  are  perpen- 
dicular to  i/,  but  the  method  of  construction  is  not  changed  when 
the  conditions  are  otherwise. 

Case  2.    To  find  the  true  size  of  the  intersection. 

Revolve  the  plane  S  in  Fig.  149  about  S-s'  into  V.  The  points 
X,  r,  Z,  and  W  will  fall  at  x^,  y^,  z^\  and  w^  respectively  (see 
Section  91),  and  x^'-y^'-z^^-w^  will  represent  the  true  size  of  the 
intersection. 


162  DESCRIPTIVE  GEOMETRY 

Case  3.    To  develop  the  surface. 

Analysis.  Take  the  plane  of  the  face  A-B-D,  Fig.  149,  as  the 
plane  of  development.  Starting  with  this  face  in  the  plane,  roll  the 
pyramid  from  left  to  right,  bringing  first  the  face  A-D-E,  second 
the  face  A-E-F^  and  finally  the  face  A-B-F^  all  into  coincidence 
with  the  original  plane. 

Construction.  Fig.  150  shows  how  these  faces  will  appear  after 
development. 

The  triangle  A-B-D,  Fig.  150,  in  which  A-B,  B-D,  and  A-D 
are  equal  respectively  to  A-B,*  h-d^^  and  A-D  of  Fig.  149,  repre- 
sents the  face  A-B^D  of  Fig.  149.  Then  since  the  face  A-D-E^ 
Fig.  149,  is  revolved  about  the  edge  A-D  as  an  axis,  it  will 
in  development  take  the  position  A-D-E,  Fig.  150,  where  A-I), 
D-E,  and  A-E  are  equal  respectively  to  A-D,  dj-e,,  and  A-E 
of  Fig.  149. 

By  a  similar  process  the  remainder  of  the  figure  may  be  con- 
structed. 

The  vertices  of  the  polygon  of  intersection  will  in  development 
take  the  positions  X,  r,  Z,  TV,  and  X'  in  Fig.  150,  where  A-X, 
A-Y,  A-Z,  A-W,  and  A-X'  are  made  equal  respectively  to  ^-X,f 
A-Y,  A-Z,  A-W,  and  A-X  of  Fig.  149. 

The  surface  A-B-D-E-F-B' ,  Fig.  150,  represents  the  develop- 
ment of  the  surface  of  the  pyramid.  The  broken  line  X-  Y-Z-  W-X' 
represents  the  development  of  the  line  of  intersection.  The  sur- 
face X-B-D-E-F-B'-X'-W-Z-Y  represents  the  development  of 
that  portion  of  the  surface  of  the  pyramid  below  the  plane  >S^. 

In  cases  of  oblique  and  irregular  pyramids  the  edges  of  the 
pyramid  will  not  be  equal,  neither  will  the  sides  of  the  base  neces- 
sarily be  equal,  as  in  the  case  just  considered. 

374.  Problem  246.  Given  an  oblique  pentagonal  pyramid,  and 
given  a  cutting  plane  perpendicular  to  V  hut  oblique  to  H ;  required 
to  find  the  intersection,  the  true  size  of  this  intersection,  and  the 
development  of  the  surface  of  the  pyramid. 

*  In  Fig.  149  the  distances  A-B,  J.-D,  A-E,  etc.,  are  not  equal  to  a'-b\  a'-d\ 
a'-e',  etc. 

t  In  Fig.  149  the  distances  A-X,  A-Y,  A-Z,  etc.,  are  not  equal  to  a'-x\  a'-y\ 
a'-z\  etc. 


INTEESECTION  OF  SURFACES  BY  PLANES         163 

375.  Problem  247.  Given  a  triangular  pyramid,  and  given  a  cut- 
ting plane  parallel  to  G—L  hut  oblique  to  Hand  V ;  required  to  find 
the  intersection,  the  true  size  of  this  intersection,  and  the  development 
of  the  surface  of  the  pyramid. 

376.  Problem  248.  Given  an  oblique  hexagonal  pyramid,  and  given 
a  cutting  plane  parallel  to  H ;  required  to  find  the  intersection,  the 
true  size  of  this  intersection,  and  the  development  of  the  surface  of 
the  pyramid. 

377.  Problem  249.  To  find  the  intersection  of  a  right  circular 
cylinder  whose  axis  is  perpendicular  to  H  by  a  plane  ;  to  draw  a 
rectilinear  tangent  to  the  curve  of  intersection  ;  to  find  the  true  size 
of  the  intersection,  and  to  develop  the  surface  of  the  cylinder. 

Case  1.    To  find  the  intersection. 

Analysis  1.  Pass  a  series  of  auxiliary  planes  parallel  to  the  axis 
of  the  cylinder,  cutting  from  the  cylinder  elements  and  cutting 
from  the  plane  straight  lines. 

Analysis  2.  Pass  a  series  of  auxiliary  planes  perpendicular  to 
the  axis,  cutting  from  the  cylinder  circles  and  cutting  from  the 
plane  straight  lines. 

Construction.  Let  the  cylinder  whose  axis  is  assumed  perpen- 
dicular to  If  be  represented  as  in  Fig.  151,  and  let  S  represent 
the  cutting  plane. 

Pass  an  auxiliary  plane  T  through  the  axis  and  perpendicular 
to  S-Sf.  T  cuts  the  cylinder  in  two  elements,  A-B  and  D-E,  and 
intersects  S  in  a  straight  line  F-G.  F-G  crosses  A-B  and  D-E  at 
K  and  E  respectively,  two  points  of  the  required  curve  of  inter- 
section. 

Owing  to  the  position  which  the  plane  T  occupies  with  reference 
to  S-Sj,  the  points  K  and  E  must  be  respectively  the  highest  and 
the  lowest  points  of  the  curve  of  intersection. 

Pass  another  auxiliary  plane  U  perpendicular  to  H  and  parallel 
to  S-Sj.  The  plane  U  cuts  the  cylinder  in  two  elements,  L-M  and 
N-0,  and  intersects  the  plane  S  in  the  straight  line  P-O-M.  The 
line  P-O-M  crosses  L-M  and  N-0  at  M  and  0  respectively,  two 
more  points  of  the  required  curve. 

By  passing  other  auxiliary  planes  parallel  to  U  we  may  obtain 
a  sufficient  number  of  points  to  locate  the  curve. 


164 


DESCRIPTIVE   GEOMETRY 


Case  2.  To  draw  a  rectilinear  tangent  to  the  curve  of  intersection. 
Analysis.  A  rectilinear  tangent  to  the  curve  of  intersection  will 
be  tangent  to  the  surface  of  the  cylinder  at  the  given  point,  and  will 
therefore  lie  in  a  plane  tangent  to  the  cylinder  along  the  element 
passing  through  this  point.  The  rectilinear  tangent  will  also  be  in 
the  plane  of  the  curve  which  is  the  cutting  plane.  The  required 
line  will  therefore  be  at  the  intersection  of  these  two  planes. 

Construction.    Let  it  be  required  to  draw  a  rectilinear  tangent  to 
the  curve  of  intersection  at  the  point  0,  Fig.  151. 

The  plane  tangent  to  the  cyl- 
inder along  the  element  through 
0  is  the  plane  W.  The  intersection 
of  this  plane  with  S  is  Q-O-B^ 
the  required  tangent. 

A  rectilinear  tangent  like  this 


Fig.  161 


Fig.  152 

can  be  drawn  at  any  point  on  the 
curve  before  the  curve  itself  has 
been  drawn. 

Since  a  rectilinear  tangent  to 
a  curve  shows  the  direction  of  the  curve  at  the  point  of  tangency, 
such  tangents,  when  determined  in  sufficient  number,  are  helpful 
in  drawing  the  curve  through  the  points  already  located. 
Case  3.  To  find  the  true  size  of  the  intersection. 
Analysis  and  Construction.  See  Fig.  151.  Revolve  the  plane  S^ 
which  is  the  plane  of  the  curve  of  intersection,  about  S-s'  into  V. 
Any  point,  as  Jf,  will  fall  at  m^,  and  other  points  may  be  found  in 
the  same  way.  The  curve  determined  by  the  revolved  positions  of 
these  points  is  the  required  curve  of  intersection. 

The  tangent  Q-0  will,  when  revolved,  take  the  position  cf'-o^'-r^ 
and  be  tangent  to  the  revolved  position  of  the  curve  of  intersection. 


INTERSECTION   OF   SURFACES  BY  PLANES         165 

Case  4.    To  develop  the  surface. 

Aiialysis  and  Construction.  Take  the  plane  W^  Fig.  151,  as  the 
plane  of  development.  Starting  with  the  element  of  tangency 
N-0  in  this  plane,  roll  the  cylinder  along  the  plane  away  from  F, 
bringing  the  successive  elements  into  contact  with  W. 

Since  the  plane  of  the  base  of  the  cylinder  is  perpendicular  to 
the  elements,  the  curve  of  the  base  will  roll  out  into  a  straight 
line  perpendicular  to  the  elements. 

The  character  of  the  development  is  shown  in  Fig.  152,  where 
N-0  represents  the  element  of  tangency,  and  where  N-N'  drawn 
perpendicular  to  N-0  and  made  equal  to  the  rectification  of  the 
curve  of  the  circular  base  of  the  cylinder  represents  the  develop- 
ment of  the  base. 

The  elements  through  A,  Z,  i),  etc.,  in  Fig.  151,  will  take  the 
positions  A-K,  L-M,  D-E^  etc.,  in  Fig.  152,  where  N-A^  A-L, 
L-D,  etc.,  are  made  equal  respectively  to  the  rectified  arcs  w,-a,, 
a-lf^  If-df,  etc.,  of  Fig.  151. 

The  points  0,  K^  Jf,  E,  0,  Fig.  151,  which  are  situated  upon 
the  elements  just  named,  will  in  development  take  the  positions 
0,  K,  M,  E,  0'  in  Fig.  152,  where  N-0,  A-K,  L-M,  etc.,  are 
made  equal  respectively  to  the  distances  of  the  same  name  in 
Fig.  151. 

The  development  of  the  curve  of  intersection  is  represented  by 
0-K-M-E-O'. 

The  tangent  0-Q  of  Fig.  151  will  in  development  take  the 
position  0-Q  in  Fig.  152,  where  N-Q  is  made  equal  to  n,-q,  of 
Fig.  151. 

378.  Problem  250.  Given  a  right  circular  cylinder  whose  axis  is 
perpendicular  to  H,  and  given  a  cutting  plane;  required  to  find  the 
intersection  hy  passing  the  auxiliary  planes  parallel  to  H. 

379.  Problem  251.  G-iven  a  right  circular  cylinder  whose  axis  is 
perpendicidar  to  H,  and  given  a  cutting  plane  which  is  parallel  to 
G-L  hut  oblique  to  H  and  V;  required  to  find  the  intersection,  to 
draw  a  rectilinear  tangent  to  the  curve  of  intersection,  to  find  the 
true  size  of  the  intersection,  and  to  develop  the  surface  of  the  cylinder. 

380.  Problem  252.  Solve  the  above  problem  when  the  cutting  plane 
is  perpendicular  to  V  but  oblique  to  H. 


166  DESCRIPTIVE  CxEOMETRY 

381.  Problem  253.  To  find  the  intersection  of  an  oblique  cylinder 
by  any  plaiie^  to  draw  a  rectilinear  tangent  to  the  curve  of  intersec- 
tion^ to  show  the  true  size  of  the  intersection,  and  to  develop  the 
surface  of  the  cylinder. 

Case  1.    To  find  the  intersection. 

Analysis.  Pass  the  auxiliary  cutting  planes  parallel  to  the  axis 
of  the  cylinder  and  perpendicular  to  H. 

Construction.  Let  the  cylinder  be  represented  as  in  Fig.  153, 
and  let  S^  in  this  case  assumed  perpendicular  to  the  elements  of 
the  cylinder,  represent  the  given  cutting  plane. 

Pass  an  auxiliary  plane  T  through  the  axis  and  perpendicular 
to  H.  This  plane  cuts  the  cylinder  in  two  elements,  A-B  and 
D-E,  and  intersects  the  plane  S  in  F-G.  F-G  crosses  A-B  and 
D-E  at  B  and  E  respectively,  two  points  in  the  required  curve 
of  intersection. 

Owing  to  the  position  of  the  plane  T  with  reference  to  S-s^^ 
the  points  B  and  E  will  be  respectively  the  lowest  and  the  highest 
points  of  the  curve  of  intersection. 

Pass  another  auxiliary  plane  U  parallel  to  T.  This  plane  cuts 
the  cylinder  in  two  elements,  K-L  and  M-N,  and  intersects  the 
plane  S  in  0-P,  where  0-P  is  necessarily  parallel  to  F-G.  0-P 
crosses  K-L  and  M-N  at  L  and  N  respectively,  two  more  points 
in  the  required  curve. 

By  passing  other  auxiliary  planes  parallel  to  T  we  may  obtain 
a  sufficient  number  of  points  to  locate  the  required  curve. 

This  method  of  construction  applies  equally  well  when  the 
given  plane  S  is  oblique  to  the  elements  of  the  cylinder. 

Case  2.    To  draw  a  rectilinear  tangent  to  the  curve  of  intersection. 

Analysis.    See  Problem  249,  Case  2. 

Construction.  See  Fig.  153.  Let  it  be  required  to  draw  a  recti- 
linear tangent  to  the  curve  of  intersection  at  the  point  L.  Draw 
the  plane  W  tangent  to  the  cylinder  along  the  element  K-L  (see 
Section  302).  The  intersection  of  the  planes  1^^  and  S  is  Q-L-B, 
the  required  tangent. 

Case  3.    To  find  the  true  size  of  the  intersection. 

Analysis  and  Construction.  See  Fig.  153.  Revolve  the  plane  S, 
which  is  the  plane  of  the  curve,  about  S-s,  into  H.   Any  point,  as  B, 


INTEKSECTIOIN'  OF   SURFACES  BY  PLANES 


167 


will  fall  at  5^,  and  other  points  may  be  found  in  the  same  way.    The 
curve  traced  through  the  points  thus  found  is  the  curve  required. 
The  tangent  E-L  will,  when  revolved,  take  the  position  r-ljj 
and  be  tangent  to  the  revolved  position  of  the  curve. 
Case  4.    To  develop  the  surface. 

Analysis.    When  the  surface  of  a  cylinder  is  rolled  out  upon  a 
plane  tangent  to  the  surface,  any  section  of  the  surface  made  by 

a  plane  perpendicular 
yii  to  the  elements  will 
roll  out  into  a  straight 
line  perpendicular  to 
the  original  element  of 
tangency.  The  work 
of  development  can 
therefore  be  much  sim- 
plified by  making  a 


Fig.  153 


Fig.  154 


right  section  of  the  surface  and  using  this  line  in  development  as 
a  line  of  reference. 

Construction.  Take  the  plane  W,  Fig.  153,  as  the  plane  of 
development,  and  take  the  point  of  observation  to  the  left  of 
the  plane.  Starting  with  the  element  of  tangency  A-i,  in  this 
plane,  roll  the  cylinder  along  this  plane  toward  F,  bringing  the 
successive  elements  of  ^he  surface  into  contact  with  W. 

Since  the  plane  S  is  perpendicular  to  the  elements,  the  curve 
L-B-X-E-N-L  will  roll  out  into  a  straight  line  perpendicular  to 
the  elements. 

The  character  of  the  development  is  shown  in  Fig.  154, 
where  K-L  represents  the  element  of  tangency  and  where  L-L\ 


168  DESCRIPTIVE  GEOMETRY 

perpendicular  to  K-L  and  equal  in  length  to  the  rectification  of 
the  true  size  of  the  right  section  of  the  surface  made  by  tlie  plane 
S^  represents  the  development  of  this  section. 

The  points  X,  B,  X,  E,  etc.,  of  Fig.  153  will,  in  Fig.  154,  take  the 
positions  X,  B^  X,  E^  etc.,  where  the  distances  L-B^  Z-X,  L-E^  etc., 
are  made  equal  to  the  rectified  arcs  Iji-hji^  hi~^^n~^in  ^H~^H~^H—^Ht 
etc.,  of  Fig.  153. 

The  elements  through  the  points  i,  i?,  E^  X,  etc.,  Fig.  153,  will 
in  development,  since  they  are  perpendicular  to  the  plane  of  the 
section,  take  the  positions  L-K^  B-A^  E-D^  etc..  Fig.  154. 

The  points  K^  A,  D,  etc..  Fig.  153,  which  are  situated  upon  the 
elements  just  named  and  also  situated  in  the  base  of  the  surface, 
will  in  development  take  the  positions  if,  A,  Z>,  etc.,  Fig.  154, 
where  the  distances  L-K,  B-A,  E-D,  etc.,  are  made  equal  to  the 
actual  distances  of  the  same  name  in  Fig.  153. 

The  development  of  the  curve  of  the  base  of  the  surface  is  then 
K-A-D-M-K',  Fig.  154. 

The  development  of  that  portion  of  the  surface  of  the  cylinder 
between  J£  and  the  plane  *S'  is  the  surface  K-A-D-K'-L'-L^ 
Fig.  154. 

The  tangent  r-k^^  tangent  to  the  base  of  the  cylinder  at  the 
point  A:,,  Fig.  153,  will  in  development  take  the  position  B-K, 
Fig.  154,  where  L-R  is  made  equal  to  Ijj-r,  of  Fig.  153. 

382.  Problem  254.  G-iven  an  oblique  cylinder  with  its  axis  in  a 
profile  plane,  and  given  a  cutting  plane  parallel  to  G-L  hut  oblique  to 
H  and  V;  required  to  find  the  intersection,  to  draw  a  rectilinear 
tangent  to  the  curve  of  intersection,  to  find  the  true  size  of  the 
intersection,  and  to  develop  the  surface  of  the  cylinder. 

383.  Problem  255.  Given  a  right  circular  cylinder  with  its  axis 
parallel  to  G-L,  and  given  a  cutting  plane  oblique  to  H  and  V; 
required  to  find  the  intersection,  to  find  the  true  size  of  the  intersec- 
tion, and  to  develop  the  surface  of  the  cylinder. 

384.  Problem  256.  Given  an  oblique  cylinder  and  a  cutting  plane 
oblique  to  the  elements  of  the  cylinder;  required  to  find  the  intersec- 
tion, to  draw  a  rectilinear  tangent  to  the  curve  of  intersection,  to  fijid 
the  true  size  of  the  intersection,  and  to  develop  the  surface  of  the 
cylinder. 


INTERSECTION  OF  SURFACES  BY  PLANES 


169 


385.  Problem  257.  To  find  the  intersection  of  a  right  circular  cone 
hy  an  oblique  plane,  to  draw  a  rectilinear  tangent  to  the  curve  of 
intersection,  to  find  the  t7'ue  size  of  the  intersection^  and  to  develop 
the  surface  of  the  cone. 

Case  1.    To  find  the  intersection. 

Analysis  1.  Pass  the  auxiliary  planes  through  the  axis  of 
the  cone. 

Analysis  2.    Pass  the  auxiliary  planes  perpendicular  to  the  axis. 


Construction.  Let 
the  cone  be  represented  as 
in  Fig.  155,  and  let  S  rep- 
resent the  cutting  plane. 

Following  Analysis  1,  let 
T  represent  one  of  the  aux- 
iliary planes.  This  plane 
intersects  the  surface  of 
the  cone  in  two  elements, 
B-A  and  B-D,  and  inter- 
sects S  in  the  line  E-F, 
E-F  crosses  B-A  and  B-D  at  G  and  K  respectively,  two  points  in 
the  required  curve  of  intersection. 

Another  auxiliary  plane  U  parallel  to  V  will  intersect  the  sur- 
face in  the  two  extreme  elements  B-L  and  B-M  and  will  locate 
the  two  points  N  and  0. 

Another  auxiliary  plane  W,  parallel  to  iS'-s,,  will  intersect  the 
surface  of  the  cone  in  the  two  elements  B-P  and  B-Q  and  will 
locate  the  two  points  B  and  X 


Fig.  155 


170  DESCEIPTIVE  GEOMETRY 

Case  2.    To  draw  a  rectilinear  tangent  to  the  curve  of  intersection. 

Analysis.   See  Problem  249,  Case  2. 

Construction.  See  Fig.  155.  Let  it  be  required  to  draw  a  recti- 
linear tangent  to  the  curve  at  the  point  R. 

Draw  the  plane  Z  tangent  to  the  cone  along  the  element  B-R-P. 
The  intersection  of  the  planes  Z  and  S  is  the  required  tangent. 

This  intersection  pierces  JT  at  «/,,  pierces  V  at  <?',  pierces  the 
plane  of  the  base  at  /,  and  must  also  pass  through  R.  Therefore 
Y—R—C—I  is  the  tangent  sought. 

Case  3.    To  find  the  true  size  of  the  intersection. 

Analysis  and  Construction.  See  Fig.  155.  Revolve  the  plane  S 
about  S-s,  into  H.  After  revolution  any  point,  as  N,  will  fall  at 
n^j^  and  other  points  may  be  found  in  the  same  way.* 

The  true  size  of  the  curve  of  intersection  is  represented  by 
''^h-9h-^h—^h~^h~'^h  ' 

Case  4.    To  develop  the  surface. 

Analysis  and  Construction.  Take  the  plane  Z,  Fig.  155,  as  the 
plane  of  development.  Starting  with  the  element  of  tangency, 
R-P,  in  this  plane,  roll  the  cone  along  this  plane  toward  V,  bring- 
ing the  successive  elements  into  contact  with  the  plane  Z. 

Since  the  cone  is  one  of  revolution  and  the  plane  of  the  base  is 
perpendicular  to  the  axis,  the  curve  of  the  circular  base  will  roll 
out  into  the  arc  of  a  circle  whose  radius  is  equal  to  the  slant 
height  of  the  cone. 

The  character  of  the  development  is  shown  in  Fig.  156,  where 
B  represents  the  vertex  of  the  cone.  The  arc  P-D-M-Q-A-L-P\ 
with  center  at  B^  with  radius  equal  to  the  slant  height  of  the  cone, 
and  equal  in  length  to  the  rectification  of  the  curve  of  the  circular 
base  of  the  cone,  represents  the  development  of  the  base. 

To  represent  the  elements  in  development,  make  P-i>,  P-M.,  P-Q-, 
etc..  Fig.  156,  such  that  the  rectification  of  their  arcs  shall  be  equal 
to  the  rectification  of  the  corresponding  arcs  of  the  same  name  upon 
the  base'of  the  original  cone,  and  connect  these  points  with  B. 

To  develop  the  curve  of  intersection,  lay  off  from  B  on  the  ele- 
ments now  in  development,  Fig.  156,  the  distances  B-R.,  B-K.,  B-0, 

*  In  the  revolution  of  these  points,  distances  from  the  axis  S-s,  have  all  been 
diminished  by  a  constant  quantity. 


INTERSECTION  OF   SURFACES  BY  PLANES         171 

etc.,  equal  respectively  to  the  distances  of  the  points  R,  K^  0,X,  etc., 
on  the  surface  of  the  cone,  from  the  vertex,  and  trace  the  curve 
E-K-0-X-G-N-R'  through  these  points. 

In  Fig.  155  the  true  distance  from  ^  to  A^  is  expressed  by  the 
distance  h'-n\  since  the  element  B-L  is  parallel  to  V.  The  dis- 
tance B-0  is  expressed  by  h'—o'  for  the  same  reason. 

The  distance  from  B  to  any  other  point  on  the  surface  may  be 
found  by  revolving  the  cone  about  its  axis  until  the  element  on 
which  the  point  stands  is  parallel  to  V.  The  vertical  projection  of  the 
required  distance  in  this  position  will  be  equal  to  the  distance  itself. 

For  example,  suppose  it  is  required  to  find  the  distance  B-X  on 
the  element  B-Q.  After  revolution  the  element  B-Q  will  take  the 
position  B-M^  vertically  projected  at  h'-m'.  The  point  x'  will  take 
the  position  x"  and  h'-x'^  will  be  the  measure  of  the  distance  B-X. 

The  tangent  R-I  of  Fig.  155  will  in  development  take  the  posi- 
tion ^-/,  Fig.  156,  where  P-I  is  drawn  tangent  to  P-D-M-A-F' 
at  P,  and  where  P-I  is  made  equal  to  p,-i,,  Fig.  155. 

386.  Problem  258.  Given  a  right  circular  cone  with  axis  perpen- 
dicular to  H,  and  given  an  intersecting  plane  oblique  to  H  and  V; 
required  to  find  the  intersection  hy  passing  auxiliary  plaiies  parallel 
to  H. 

387.  Problem  259.  Given  a  right  circular  cone  with  axis  perpen- 
dicular to  i/,  and  given  an  intersecting  plane  parallel  to  H;  required 
to  determine  the  character  of  the  intersection. 

388.  Problem  260.  Given  a  right  circular  cone  with  axis  perpen- 
dicular to  H^  and  given  an  intersecting  plane  passing  through  the 
axis;  required  to  determine  the  character  of  the  intersection. 

389.  Problem  261.  Given  a  right  circular  cone  with  axis  perpen- 
dicular to  H^  and  given  an  intersecting  plane  making  a  smaller  angle 
with  H  than  the  elements  of  the  cone;  required  to  determine  the 
character  of  the  intersection. 

390.  Problem  262.  Solve  Problem  261  when  the  intersecting  plane 
makes  the  same  angle  tvith  H  as  the  elements  of  the  cone. 

391.  Problem  263.  Solve  Problem  261  ivhen  the  intersecting  plane 
makes  a  greater  angle  with  H  than  the  elements  of  the  cone. 

392.  Conic  Sections.  By  an  examination  of  the  results  obtained 
in  Problems  259  and  260  it  will  be  observed  that  when  the  surface 


172  DESCRIPTIVE  GEOMETRY 

of  a  right  circular  cone  is  cut  by  a  plane  perpendicular  to  the  axis, 
the  curve  of  intersection  is  the  circumference  of  a  circle ;  and  when 
cut  by  a  plane  containing  the  axis,  the  intersections  are  elements. 

By  examination  of  the  results  obtained  in  Problems  261-263, 
and  by  reference  to  treatises  on  solid  and  analytic  geometry,  it 
may  be  shown  that  when  a  right  circular  cone  with  axis  perpen- 
dicular to  H  is  cut  by  a  plane  making  a  smaller  angle  with  H  than 
the  elements  of  the  cone,  the  curve  of  intersection  is  an  ellipse; 
when  cut  by  a  plane  making  the  same  angle  with  H  as  the  elements 
of  the  cone,  the  curve  of  intersection  is  a  parabola;  and  when  cut 
by  a  plane  making  a  greater  angle  with  H  than  the  elements  of 
the  cone,  the  curve  of  intersection  is  a  hyperhola. 

393.  Problem  264.  To  find  the  intersection  of  any  cone  by  a  plane, 
to  draw  a  rectilinear  tangent  to  the  curve  of  intersection,  and  to  find 
the  true  size  of  the  intersection. 

Case  1.    To  find  the  intersection. 

Analysis  and  Construction.  Let  the  cone  be  represented  as  in 
Fig.  157,  and  let  S  represent  the  intersecting  plane. 

Pass  the  auxiliary  planes  through  the  vertex  B  and  perpendicular 
to  H.  Then  all  the  auxiliary  planes  will  intersect  in  a  common 
straight  line  through  the  vertex  and  perpendicular  to  H.  This 
straight  line  will  intersect  S  2X  A  (see  Section  151),  which  is  there- 
fore a  point  common  to  all  the  lines  cut  from  S  by  the  auxiliary 
planes. 

Let  T  represent  one  of  the  auxiliary  planes.  This  plane  inter- 
sects the  surface  of  the  cone  in  two  elements,  B-D  and  B-E,  and 
intersects  the  plane  S  in  the  line  F-A,  passing  through  the  point 
A  previously  located.  F-A  crosses  B-D  and  B-E  at  K  and  L 
respectively,  two  points  in  the  required  curve  of  intersection. 

Another  auxiliary  plane  U  will  intersect  the  surface  of  the  cone 
in  the  elements  B-M  and  B-N,  and  will  intersect  the  plane  S  in 
the  line  I- A.  I- A  crosses  B-M  oxidi  B-N  dX  0  and  P  respectively, 
two  more  points  in  the  required  curve. 

Case  2.    To  draw  a  rectilinear  tangent  to  the  curve  of  intersection. 

Analysis.   See  Problem  249,  Case  2. 

Construction.  See  Fig.  157.  Let  it  be  required  to  draw  a  recti- 
linear tangent  to  the  curve  at  the  point  X.    Draw  the  horizontal 


INTERSECTION  OF   SURFACES   BY  PLANES 


173 


trace  of  the  plane  W  tangent  to  the  cone  along  the  element  B-L-E 
(see  Section  317).  The  intersection  of  the  planes  W  and  S  is  the 
required  tangent.  This  intersection  pierces  H  at  q^  and  must  also 
pass  through  L.    Therefore  Q-L  is  the  required  tangent. 

Case  3.    To  find  the  true  size  of  the  intersection. 

Analysis  and  Construction.  See  Fig.  157.  Revolve  the  plane  S 
about  S-Sj  into  H. 

After  revolution  any  point,  as  0,  will  fall  at  o^^,  and  other  points 
may  be  found  in  the  same  way. 

Or  revolve  A^  the  point  common  to  all  the  lines  cut  from  S  by 
the  auxiliary  planes,  about  S-s^  into  H.    Then  all  of  the  lines  cut 


from  S  by  the   auxiliary  planes  will  in  revolved  position  pass 
through  the  revolved  position  of  this  point. 

For  example,  A-F  will  in  revolved  position  fall  at  a^-f^.  The 
points  K  and  L  upon  this  line  A-F  will  move  in  planes  perpen- 
dicular to  S-s^  and  will  therefore  fall  at  hj^  and  l^j  respectively, 
upon  lines  through  h^  and  l^  perpendicular  to  S-s^. 


174  DESCRIPTIVE  GEOMETRY 

The  true  size  of  the  curve  of  intersection   is  represented  by 

The  tangent  Q-L  will  when  revolved  take  the  position  q-ln, 
the  point  q^  remaining  stationary. 

The  development  of  an  oblique  cone  requires  knowledge  of  a  prob- 
lem not  yet  explained,  and  will  be  taken  up  later  (see  Section  421). 

394.  Problem  265.  G-iven  an  oblique  cone  with  base  on  F,  and  given 
a  cutting  plane  oblique  to  G—L;  required  to  find  the  intersectio7i,  to 
draiv  a  rectilinear  tangent  to  the  curve  of  intersection^  and  to  find 
the  true  size  of.  the  intersection. 

395.  Problem  266.  Given  an  oblique  cone  tvith  axis  in  a  profile 
plane,  and  given  an  intersecting  plane  parallel  to  G—L  but  oblique  to 
H  and  V;  required  to  find  the  intersection,  to  draw  a  rectilinear 
tangent  to  the  curve  of  intersection,  and  to  find  the  true  size  of  the 
intersection. 

396.  Problem  267.  Given  a  right  circular  cone  with  axis  parallel 
to  G-L,  and  given  a  chitting  p>lane  oblique  to  G-L ;  required  to  find 
the  iyitersection,  to  draw  a  rectilinear  tangent  to  the  curve  of  intersec- 
tion, and  to  find  the  true  size  of  the  intersection. 

397.  Problem  268.  To  find  the  intersection  of  any  surface  of  revo- 
lution by  a  pla7ie,  to  draw  a  rectilinear  tangent  to  the  curve  of  inter- 
section, and  to  find  the  true  size  of  the,  intersection. 

Case  1.    To  find  the  i7itersection. 

Analysis.  Pass  the  auxiliary  planes  perpendicular  to  the  axis 
(see  Section  284). 

Construction.  Let.  the  surface  of  revolution  be  represented  as  in 
Fig.  158,  where  A-B  is  the  axis  and  where  >S'  is  the  cutting  plane. 

Let  T  represent  one  of  the  auxiliary  planes  perpendicular  to 
A-B.  This  plane  cuts  the  given  surface  in  the  circumference  of 
a  circle  whose  vertical  projection  is  d'-e'  and  whose  horizontal 
projection  is  d-g-e-h,.  This  same  plane  intersects  S  in  the 
straight  line  6r-/i-i^  parallel  to  S-s^.  G-K-F  crosses  the  circum- 
ference IJ-G-E-K  at  G  and  K,  two  points  of  the  required  curve 
of  intersection. 

Pass  a  meridian  plane  U  perpendicular  to  S-s^,  cutting  from 
the  surface  a  meridian  curve  and  cutting  S  in  a  straight  line 
JI-L  which  intersects  the  axis  at  M. 


INTERSECTION  OF   SURFACES  BY  PLANES 


to 


To  find  the  points  in  which  Af-L  intersects  the  meridian  curve, 
revolve  the  plane  U  about  the  axis  A-B  until  it  is  parallel  to  V. 
The  vertical  projection  of  the  meridian  curve  will  then  be  identical 
with  the  vertical  projection  of  the  surface,  and  the  vertical  projec- 
tion of  M-L  will  fall  at  m'-l",  M  remaining  stationary.  The  points 
n"  and  o'\  the  points  in  which  the  line  m'-V  crosses  the  vertical 
projection  of  the  meridian  curve,  are  the  vertical  projections  of 


Fig.  158 


the  revolved  positions  of  the  lowest  and  the  highest  points  of  the 
required  curve  of  intersection.  The  projections  of  N  in  true  posi- 
tion will  fall  at  n^  and  ?^',  and  the  projections  of  0  in  true  position 
will  fall  at  0,  and  o'. 

A  meridian  plane  W,  parallel  to  F,  will  cut  from  the  surface  a 
meridian  curve  whose  vertical  projection  is  the  vertical  projection 


176  DESCEIPTIVE  GEOMETRY 

of  the  surface,  and  will  cut  from  S  a  straight  line  whose  vertical 
projection  is  p'-m\  parallel  to  S-s',  By  this  plane  W  two  more 
points,  Q  and  R^  are  located. 

Case  2.    To  draw  a  rectilinear  tangent  to  the  curve. 

By  analyses  previously  given,  a  rectilinear  tangent  to  the  curve, 
cut  from  any  surface  by  a  plane,  may  be  drawn  whenever  a  plane 
can  be  drawn  tangent  to  the  surface  at  the  point  of  tangency. 

Case  3.    To  find  the  true  size  of  the  intersection. 

Analysis  and  Constructio7i.  See  Fig.  158.  Revolve  the  plane 
S^  the  plane  of  the  curve,  about  S-s^  into  H.  Any  point  of  the 
curve,  as  0,  will  fall  at  o^^,  and  other  points  may  be  found  in  the 
same  way.* 

Note.  In  the  following  six  problems  apply  the  principles  suggested  by  the 
analysis  in  Problem  268,  Case  1. 

398.  Problem  269.  Find  the  intersection  of  a  cylinder  of  revolution 
by  an  oblique  plane. 

399.  Problem  270.  Find  the  intersection  of  a  cone  of  revolution 
by  an  oblique  plane. 

400.  Problem  271.  Find  the  intersection  of  a  sphere  by  a  plane 
oblique  to  G—L. 

401.  Problem  272.  Find  the  intersection  of  an  ellipsoid  of  revolu- 
tion by  an  oblique  plane. 

402.  Problem  273.  Find  the  intersection  of  a  paraboloid  of  revo- 
lution by  an  oblique  plane. 

403.  Problem  274.  Find  the  intersection  of  a  hyperboloid  of  revo- 
lution by  an  oblique  plane. 

*  In  this  revolution  the  radii  have  all  been  shortened  by  a  constant  quantity  in 
order  to  reduce  the  space  occupied  by  the  drawing. 


CHAPTER  XYII 

INTERSECTION  OF  SURFACES  BY  SURFACES 

404.  General  Instructions.  Pass  a  series  of  auxiliary  planes  so  as 
to  cut  lines  from  the  two  surfaces.  The  points  in  which  the  lines 
of  one  surface  intersect  the  lines  of  the  other  are  necessarily  in  both 
surfaces  and  therefore  in  their  line  of  intersection. 

Pass  the  auxiliary  planes  in  such  a  way  as  not  only  to  cut  from 
the  surfaces  the  simplest  lines  but  also  in  such  a  way  as  to  make 
the  work  of  construction  as  simple  as  possible. 

If  the  two  surfaces  are  of  revolution,  with  intersecting  axes,  it 
will  often  be  found  convenient  to  pass  a  series  of  auxiliary  cutting 
spheres  with  centers  at  the  intersection  of  the  axes.  A  sphere  of 
this  character  will  cut  from  the  two  surfaces  circumferences  of 
circles  which,  lying  on  the  surface  of  the  sphere,  will  as  a  rule  inter- 
sect. These  points  of  intersection  must  lie  on  the  line  in  which  the 
given  surfaces  intersect,  since  they  are  common  to  both  surfaces. 

405.  Problem  275.  To  find  the  intersection  of  a  cylinder  and  a 
cone^  to  draw  a  rectilinear  tangent  to  the  curve  of  intersection,  and 
to  develop  the  surfaces. 

Case  1.    To  find  the  intersection. 

Analysis.  Pass  the  auxiliary  planes  through  the  vertex  of  the 
cone  and  parallel  to  the  axis  of  the  cylinder. 

Construction.  Let  the  cylinder  be  represented  with  its  base  on 
H,  and  let  the  cone  be  represented  with  its  base  on  F,  as  shown  in 
Fig.  159.  The  axis  of  the  cylinder  is  represented  by  A-B  and 
the  axis  of  the  cone  is  represented  by  C-D. 

A  straight  line  through  the  vertex  of  the  cone  parallel  to  the 
axis  of  the  cylinder  will  be  common  to  all  the  auxiliary  planes. 

Through  D  draw  D-E  parallel  to  A-B  and  produce  it  to  meet 
IT  at  Cf  and  to  meet  V  at  f.  The  horizontal  traces  of  all  the 
auxiliary  planes  must  pass  through  e^  and  the  vertical  traces  of 
all  the  auxiliary  planes  must  pass  through  f. 

177 


178  DESCRIPTIVE  GEOMETRY 

Through  e^  draw  any  horizontal  trace,  as  S-s^^  cutting  the  base 
of  the  cylinder.  Through  >S'  and  /'  draw  the  vertical  trace  S-s\ 
S  is  one  of  the  auxiliary  planes  and  cuts  the  cylinder  in  two 
elements,  G—K—L  and  M—N-0 ;  and  cuts  the  cone  in  two  elements, 
P-D  and  Q-D. 

The  element  G-K-L  of  the  cylinder  crosses  the  two  elements 
of  the  cone  at  K  and  i,  two  points  of  the  required  curve  of 
intersection.  The  element  M—N-0  of  the  cylinder  crosses  the 
two  elements  of  the  cone  at  ^Y  and  0,  tAvo  more  points  of  the 
curve. 

In  this  way  we  may  pass  any  number  of  auxiliary  planes  and 
obtain  any  number  of  points  on  the  required  curve  of  intersection. 

If  it  is  desired  to  obtain  a  point  of  the  curve  upon  any  particular 
element  of  the  cylinder  or  of  the  cone,  we  have  but  to  draw  the 
traces  of  the  auxiliary  plane  through  the  points  in  which  this 
element  pierces  the  corresponding  planes  of  projection. 

It  will  be  noticed  that  since  the  horizontal  and  vertical  projec- 
tions of  the  curve  of  intersection  are  found  independently,  the 
accuracy  of  the  work  may  be  tested  by  noting  whether  the  hori- 
zontal and  vertical  projections  of  the  several  points  of  the  curve 
lie  in  straight  lines  perpendicular  to  G-L. 

When  projecting  on  7/,  that  portion  of  the  curve  which  lies  upon 
the  upper  surface  of  the  cylinder  and  also  upon  the  upper  surface 
of  the  cone,  and  between  extreme  elements,  will  be  visible  and 
should  be  so  represented. 

When  projecting  on  F,  that  portion  of  the  curve  which  lies 
upon  the  front  surface  of  the  cylinder  and  also  upon  the  front  sur- 
face of  the  cone,  and  between  extreme  elements,  will  be  visible 
and  should  be  so  represented. 

Case  2.  To  draw  a  rectilinear  tangent  to  the  curve  of  intersection 
at  a  given  point. 

Analysis.  Since  the  curve  of  intersection  lies  upon  both  sur- 
faces, the  required  tangent  must  lie  both  in  a  plane  tangent  to  the 
cone  at  the  given  point,  and  in  a  plane  tangent  to  the  cylinder  at 
the  same  point.  Therefore  draw  two  planes,  one  tangent  to  the 
cylinder  and  the  other  tangent  to  the  cone  at  the  given  point.  The 
intersection  of  these  two  planes  is  the  required  tangent. 


Fig.  169 


179 


180 


DESCRIPTIVE  GEOMETRY 


Case  3.    To  develop  the  surfaces. 

The  surface  of  the  cylinder  and  the  curve  which  lies  upon  it 
may  be  developed  by  methods  previously  given  (see  Section  377). 
The  surface  of  the  cone  and  the  same  curve  which  also  lies  upon 
this  surface  may  be  developed  by  the  method  given  in  Section  421, 
yet  to  be  explained. 

406.  Problem  276.    To  find  the  intersection  of  two   cylinders,  to 


Fig.  160 


draw  a  rectilinear  tangent  to  the  curve  of  intersection,  and  to  develop 
the  surfaces. 

Case  1.    To  find  the  intersection. 

Analysis.  Pass  the  auxiliary  planes  parallel  to  the  axes  of  the 
two  cylinders. 

Construction.  Let  the  two  cylinders  be  represented  with  their 
two  bases  on  H,  as  shown  in  Fig.  160.  The  axis  of  one  cylinder 
is  represented  by  A-B,  and  the  axis  of  the  other  cylinder  is  repre- 
sented by  C-D, 


INTEESECTION  OF  SURFACES  BY  SURFACES   181 

A  plane  through  one  of  the  axes  and  parallel  to  the  other  will 
itself  be  one  of  the  auxiliary  planes,  and  will  be  parallel  to  all  the 
other  auxiliary  planes. 

Through  any  point  on  the  axis  (7-i>,  as  i),  draw  D-E  parallel 
to  the  other  axis  A-B.  C-D  pierces  H  at  c^  and  D-E  pierces  H 
at  e,.  S-c-e-Sf  is  the  horizontal  trace  of  an  auxiliary  plane,  and 
one  to  which  all  the  other  auxiliary  planes  must  be  parallel. 

The  plane  S  cuts  the  cylinder  whose  axis  is  A-B  in  two  ele- 
ments, F-G  and  K-L,  and  cuts  the  other  cylinder  in  two  elements, 
jyr-A^and  0-G,  These  elements  cross  in  the  points  iV",  G,  i,  and 
P,  four  points  in  the  required  curve  of  intersection. 

Other  auxiliary  planes  parallel  to  S  may  be  drawn,  and  a  suffi- 
cient number  of  points  may  be  located  to  determine  the  curve  of 
intersection. 

If  it  is  desired  to  determine  a  point  upon  any  particular  element 
of  either  cylinder,  we  have  but  to  pass  the  auxiliary  plane  so  as 
to  cut  from  the  cylinder  this  element. 

When  projecting  on  ff^  that  portion  of  the  curve  of  intersection 
which  lies  on  the  upper  surfaces  of  both  cylinders  and  between 
extreme  elements  will  be  visible. 

When  projecting  on  F,  that  portion  of  the  curve  of  intersection 
which  lies  on  the  front  surfaces  of  both  cylinders  and  between 
extreme  elements  will  be  visible. 

Case  2.  To  draw  a  rectilinear  tangent  to  the  curve  of  intersection 
at  a  given  point. 

Analysis.  See  Problem  275,  Case  2.  The  required  tangent  will 
be  the  intersection  of  two  planes,  —  one  tangent  to  one  cylinder 
at  the  point  in  question,  and  the  other  tangent  to  the  other  cylinder 
at  the  same  point.  Therefore  draw  two  planes  under  these  condi- 
tions and  find  their  line  of  intersection. 

Case  3.    To  develop  the  surfaces. 

The  surfaces  of  the  cylinders  and  the  curves  of  intersection 
which  are  common  to  both  surfaces  may  be  developed  by  rules 
previously  given  (see  Problem  253,  Case  4). 

407.  Problem  277.  To  find  the  intersection  of  two  cones,  to  draw 
a  rectilinear  tangent  to  the  curve  of  intersection^  and  to  develop  the 
surfaces. 


182 


DESCKIPTIVE   GEOMETKY 


Case  1 .    To  find  the  intersection. 

Analysis.    Draw  the  auxiliary  planes  through  the  two  vertices. 

Construction.  See  Fig.  161.  Let  A-B  represent  the  axis  of  the 
first  cone,  which  is  right  circular  with  base  parallel  to  H-,  and  let 
C-D  represent  the  axis  of  the  second  cone,  whose  base  is  a  circle 
with  its  plane  perpendicular  to  V.  The  vertical  trace  of  the  plane 
of  the  base  of  the  second  cone  is  S-s\  and  the  intersection  of  this 
plane  with  the  plane  of  the  base  of  the  first  cone  is  the  line  U-G-  W. 


Fig.  161 


A  straight  line  through  the  two  vertices  will  be  common  to  all 
the  auxiliary  planes.  Therefore  through  the  two  vertices  draw 
B-D  and  produce  it  to  meet  the  plane  of  the  base  of  the  first  cone 
at  E^  a  point  common  to  all  the  lines  cut  from  this  plane  by  the 
auxiliary  planes.  This  line  B-D  pierces  S,  the  plane  of  the  base 
of  the  second  cone,  in  F^  a  point  common  to  all  the  lines  cut  from 
*S^  by  the  auxiliary  planes. 

Through  e,  draw  e^-g^  to  represent  the  horizontal  projection  of 
the  intersection  of  one  of  the  auxiliary  planes  with  the  plane  of 


INTEKSECTION  OF   SURFACES  BY  SURFACES      183 

the  base  of  the  first  cone.    This  auxiliary  plane  cuts  from  the 
first  cone  the  two  elements  B-K  and  B-L. 

The  same  auxiliary  plane  intersects  the  plane  S  in  the  line  F-G^ 
which  must  cross  the  base  of  the  second  cone  in  points  of  the 
elements  cut  from  the  second  cone  by  this  auxiliary  plane. 

To  find  these  elements,  revolve  S  about  S-s'  into  V.  The  center 
of  the  circular  base  of  the  second  cone  will  fall  at  c^,  the  point  F 
will  fall  at  /^,  and  the  point  G  will  fall  at  g^. 

The  line /^-^^-m^  represents  the  revolved  position  of  F-G^  and 
the  points  n^'  and  rrJ^  represent  the  revolved  positions  of  the  two 
points  in  which  F-G  crosses  the  base  of  the  second  cone. 

After  the  counter  revolution  M  and  N  will  take  the  positions 
(mp  m')  and  (ti^  n')  respectively,  where  m,  and  n,  are  at  the  same 
distances  from  G-L  as  M  and  N  respectively  are  from  the  axis  of 
revolution  S-s'. 

The  two  elements  cut  from  the  second  cone  by  this  auxiliary 
plane  are  then  D-M  and  D-N.  These  elements  cross  the  tw^o  ele- 
ments cut  from  the  first  cone  by  this  same  plane  at  0,  P,  Q^  and 
B^  four  points  of  the  required  curve  of  intersection. 

Other  auxiliary  planes  may  be  drawn  in  the  same  way,  and  a 
sufficient  number  of  points  may  be  determined  to  locate  the 
required  curve  of  intersection. 

That  which  has  been  said  in  previous  sections  regarding  the 
determination  of  visible  and  invisible  portions  of  the  curve  of  inter- 
section, the  construction  of  rectilinear  tangents  to  this  curve,  and 
the  development  of  the  surfaces,  may  be  said  of  these  surfaces  also. 

408.  To  determine  in  advance  the  Nature  of  the  Curves  in  which 
Cylinders  and  Cones  intersect.  The  nature  of  the  curve  of  inter- 
section will  depend  upon  the  relative  size  and  position  of  the 
intersecting  surfaces.  The  surfaces  may  intersect  so  that  a  por- 
tion of  one  will  remain  entirely  outside  of  the  other,  giving 
one  continuous  curve  of  intersection,  as  in  Fig.  159;  or  they 
may  intersect  so  as  to  have  one  distinct  curve  of  ingress  and 
another  distinct  curve  of  egress,  as  in  Fig.  160;  or  the  two  sur- 
faces may  lie  between  two  planes  to  which  the  surfaces  are  both 
tangent,  in  which  case  the  curve  of  ingress  will  be  tangent  to  the 
curve  of  egress  on  opposite  sides  of  the  surfaces,  as  in  Fig.  161. 


184  DESCRIPTIVE  GEOMETRY 

The  nature  of  the  curve  of  intersection  may  be  determined  in 
advance  by  drawing,  under  the  same  conditions  as  the  auxiliary 
planes  are  drawn,  tangent  planes  to  the  surfaces. 

If  in  Fig.  159  we  draw,  as  one  of  the  auxiliary  planes  of  this 
problem,  the  plane  T  tangent  to  the  cylinder  along  the  upper  sur- 
face, the  position  of  its  vertical  trace  T-t'  shows  that  such  a  plane 
intersects  the  cone,  and  that  a  portion  of  the  cone  remains  entirely 
outside  the  cylinder. 

The  plane  W  drawn  under  the  conditions  mentioned  above  and 
tangent  to  the  cylinder  along  the  under  surface  does  not  cut  the 
cone,  showing  that  a  portion  of  the  cylinder  remains  outside  the 
cone. 

The  curve  of  intersection  then  in  this  case  will  be  one  contin- 
uous curve,  as  shown  in  the  figure. 

If  both  the  planes  T  and  W  had  intersected  the  cone,  the  indi- 
cation would  be  that  the  cylinder  passed  through  the  cone,  giving 
two  distinct  curves  of  intersection. 

If  the  base  of  the  cone  had  fallen  wholly  within  the  two  vertical 
traces  T-t'  and  W-w',  the  indication  would  be  that  the  cone  passed 
through  the  cylinder,  giving  two  distinct  curves  of  intersection. 

This  last  condition  is  illustrated  in  Fig.  160,  where  the  cylinder 
whose  axis  is  C-D  intersects  the  cylinder  whose  axis  is  A-B  in 
two  distinct  curves.  The  base  of  the  cylinder  whose  axis  is  C-D 
falls  wholly  within  the  two  planes  T  and  U,  which  are  drawn 
under  the  same  conditions  as  the  auxiliary  planes  of  this  problem, 
and  tangent  to  the  cylinder  whose  axis  is  A-B. 

If  the  two  vertical  traces  T-t'  and  W-w'  in  Fig.  159  had  included 
and  had  been  tangent  to  the  base  of  the  cone,  the  indication  would 
be  that  the  two  surfaces  were  included  by  the  tangent  planes  bring- 
ing the  curves  of  intersection  into  a  position  of  tangency.  This 
condition  is  illustrated  in  Fig.  161,  where  the  first  cone  whose  axis 
is  A-B  intersects  the  second  cone  in  two  tangent  curves.  In 
Fig.  161  the  two  planes  whose  intersections  with  the  plane  of  the 
base  of  the  first  cone  are  U-U  and  U-JV  are  tangent  to  the  first 
cone  and  are  also  tangent  to  the  second  cone,  as  may  be  seen  from 
the  revolved  position  of  the  lines  in  which  these  tangent  planes 
intersect  S,  the  plane  of  the  base  of  the  second  cone. 


INTERSECTION  OF  SURFACES  BY  SURFACES   185 

409.  Problem  278.  Find  the  intersection  of  a  cylinder  and  a  cone 
when  the  bases  of  both  surfaces  are  on  V. 

410.  Problem  279.  Find  the  intersection  of  two  cylinders  when  the 
base  of  one  surface  is  on  H  and  the  base  of  the  other  is  on  V. 

411.  Problem  280.  Find  the  intersection  of  two  cones  when  the  bases 
of  both  surfaces  are  on  H. 

412.  Problem  281.  Find  the  intersection  of  two  cylinders  of  the 
same  diameter,  one  horizontal  and  the  other  vertical,  axes  intersecting. 

413.  Problem  282.  Find  the  intersection  of  two  cylinders  of  unequal 
diameters,  07ie  horizontal  and  the  other  vertical,  axes  intersecting.    . 

414.  Problem  283.  Find  the  intersection  of  two  cylinders  of  unequal 
diameters,  one  horizontal  and  the  other  vertical,  axes  not  intersecting. 

415.  Problem  284.  Fi7id  the  intersection  of  two  cylinders  of  un- 
equal diameters,  the  larger  vertical,  the  smaller  inclined  at  an  angle 
of  60  degrees  to  H,  axes  intersecting. 

416.  Problem  285.  To  find  the  intersection  of  a  sphere  and  a 
cylinder  and  to  draw  a  rectilinear  tangent  to  the  curve  of  intersection 
at  a  given  point  on  the  curve. 

Case  1.    To  find  the  intersection. 

Analysis  1.  Pass  the  auxiliary  planes  parallel  to  the  axis  of  the 
cylinder,  cutting  from  the  cylinder  elements  and  cutting  from 
the  sphere  circles. 

Analysis  2.  If  the  base  of  the  cylinder  is  a  circle,  we  may  pass 
the  auxiliary  planes  parallel  to  this  base,  cutting  circles  from  both 
surfaces. 

Construction.  In  Fig.  162  let  C  represent  the  center  of  the 
sphere  and  let  A-B,  passing  through  the  center  of  the  sphere, 
represent  the  axis  of  the  cylinder. 

Pass  the  auxiliary  planes  parallel  to  A-B  and  perpendicular 
to  H.  The  horizontal  trace  of  the  auxiliary  plane  containing  the 
axis  is  S-Sj.  This  plane  cuts  the  sphere  in  a  great  circle  and  cuts 
the  cylinder  in  two  elements  which  intersect  the  circumference  of 
the  circle  in  four  points  of  the  required  curve  of  intersection. 

To  find  these  points,  revolve  the  plane  S  about  S-s,  into  H. 
C  will  fall  at  Cjj  and  the  great  circle  will  fall  at  Cjj-gjj-ljj-kjj.  The 
axis  A-B  will  fall  at  a-bjj,  and  the  two  elements  in  which  the 
auxiliary  plane  intersects  the  cylinder  will  fall  at  d-e^j  and  f-g^^ 


186 


DESCRIPTIVE  GEOMETRY 


parallel  to  a-hjj.    The  points  E^  K,  G,  and  L,  projected  at  (g,,  e'), 
(kf,  k'),  (^,,  g')^  and  (l^,  I'),  are  the  four  pomts  of  intersection. 

The  auxiliary  plane  T  cuts  the  sphere  in  a  small  circle  whose 
center  is  Jf  and  whose  diameter  is  equal  to  n,-(?,.    The  same  plane 


/^ 


[h- 


I 


m 


^/  //ni/fr\ 

1 1        r^% 


Fig.  162 


T  cuts  the  cylinder  in  two  elements,  F-Q  and  Il-Z,  which  are 
parallel  to  A-B. 

By  passing  a  sufficient  number  of  auxiliary  planes  we  may  locate 
enough  points  to  determine  the  required  curve  of  intersection. 

Since  the  auxiliary  planes  U  and  W,  which  are  drawn  tangent 
to  the  cylinder,  both  intersect  the  sphere,  there  are,  according  to 
Section  408,  two  distinct  curves  of  intersection,  as  may  be  seen 
from  the  drawing. 

The  visible  and  invisible  portions  of  the  curves  may  be  deter- 
mined by  methods  previously  explained. 


intersectio:n^  of  surfaces  by  surfaces  187 

Case  2.    To  draw  a  rectilinear  tangent  to  the  curve  of  intersection. 

The  rectilinear  tangent  to  the  curve  at  any  given  point  will  be 
the  intersection  of  two  planes,  one  tangent  to  the  cylinder  at  the 
given  point  and  the  other  tangent  to  the  sphere  at  the  same  point. 

417.  Problem  286.  Find  the  intersection  of  a  sphere  and  a  cylinder 
whe7i  the  axis  of  the  cylinder  is  vertical  and  passes  through  the  center 
of  the  sphere. 

418.  Problem  287.  Find  the  intersection  of  a  sphere  and  a  cylinder 
when  the  axis  of  the  cylinder  is  horizontal  and  passes  through  the 
center  of  the  sphere. 

419.  Problem  288.  Find  the  intersection  of  a  sphere  and  a  cylinder 
when  the  axis  of  the  cylinder  does  not  pass  through  the  center  of  the 
sphere,  and  when  some  of  the  elements  of  the  cylinder  remain  wholly 
outside  the  sphere. 

420.  Problem  289.  To  find  the  intersection  of  a  hemisphere  and  a 
cone  and  to  draw  a  rectilinear  tangent  to  the  curve  of  intersection. 

Casp:  1.    To  find  the  intersection. 

Analysis  1.  Pass  the  auxiliary  planes  through  the  vertex  of  the 
cone  and  perpendicular  to  //,  cutting  elements  from  the  cone  and 
semicircles  from  the  hemisphere. 

Analysis  2.    Same  as  Analysis  2  of  Section  416. 

Constructio7i.  Let  the  cone  and  the  hemisphere  be  represented 
as  in  Fig.  163,  where  the  vertex  of  the  cone  is  taken  at  the  center 
of  the  hemisphere. 

The  plane  >S'  represents  one  of  the  auxiliary  planes,  cutting  the 
cone  in  two  elements  B-E  and  B-P,  and  cutting  the  hemisphere 
in  a  semicircle. 

To  find  the  points  in  which  these  elements  intersect  the  semi- 
circle, revolve  >S^  about  a  vertical  axis  through  B  until  it  is  parallel 
to  V.  The  two  elements  B-E  and  B-B  will  then  be  vertically 
projected  at  h'-r"  and  h'-p"  respectively,  and  the  vertical  projection 
of  the  semicircle  will  be  coincident  with  the  vertical  projection  of 
the  hemisphere.  Therefore  u"  and  q"  are  the  vertical  projections 
of  the  revolved  positions  of  two  points  of  the  required  curve  of 
intersection. 

The  auxiliary  plane  7',  parallel  to  T,  cuts  the  cone  in  two 
elements  7/- 71  and  I>-L,  and  cuts  the  hemisphere  in  a  semicircle 


188  DESCEIPTIVE  GEOMETRY 

whose  vertical  projection  coincides  with  the  vertical  projection  of 
the  hemisphere.  This  plane  will  locate  the  points  of  the  curve 
whose  vertical  projections  fall  on  the  vertical  projection  of  the 
hemisphere. 

In  the  same  way  other  planes  may  be  passed  and  a  sufficient 
number  of  points  may  be  determined  to  locate  the  curve  of 
intersection. 

When  projecting  on  H^  if  we  regard  the  hemisphere  as  hollow, 
the  whole  curve  will  be  visible. 

When  projecting  on  F,  that  portion  of  the  curve  which  lies  on 
the  front  portions  of  both  surfaces  and  between  extreme  elements 
will  be  visible. 

Case  2.    To  draw  a  rectilinear  tangent  to  the  curve  of  intersection, 

A  rectilinear  tangent  to  the  curve  of  intersection  at  any  point 
will  be  the  intersection  of  two  planes,  one  of  which  is  tangent  to 
the  hemisphere  at  the  given  point  and  the  other  tangent  to  the 
cone  at  the  same  point. 

421.  Problem  290.    To  develop  an  oblique  cone. 

Analysis.  The  intersection  of  the  surface  of  any  cone  by  the 
surface  of  a  sphere  whose  center  is  at  the  vertex  of  the  cone  is  a 
curve  whose  points  are  all  equidistant  from  'the  vertex  of  the  cone, 
because  the  curve  lies  on  the  surface  of  the  sphere  as  well  as  on 
the  surface  of  the  cone. 

For  this  reason,  in  the  development  of  the  cone,  this  particular 
curve  of  intersection  will  roll  out  into  the  arc  of  a  circle  whose 
center  is  the  vertex  of  the  cone  and  whose  radius  is  equal  to  the 
radius  of  the  sphere. 

If  upon  this  arc,  starting  from  any  point,  we  lay  off  successively 
the  actual  arc  distances  between  the  points  in  which  the  elements 
of  the  cone  cut  the  curve  of  intersection  before  development,  and 
if  we  connect  these  points  of  division  and  the  center  of  the  arc  by 
straight  lines,  these  straight  lines  will  represent  the  elements  of 
the  cone  in  development. 

We  may  now  use  these  elements  to  determine  points  in  the 
development  of  other  curves  on  the  surface  of  the  cone. 

Construction.  Let  it  be  required  to  develop  the  oblique  cone 
given  in  Fig.  163.    The  intersection  of  the  surface  of  the  cone  by 


Fig.  164 


F,r 


Q      Q 


^IS^^' 


r, 


ill! 


■u       ^' 


Fig.  165 
189 


190  DESCRIPTIVE   GEOMETRY 

the  sphere  is  already  determined.  Take  the  plane  which  is  tangent 
to  the  cone  along  the  element  B-D  as  the  plane  of  development. 

Then  in  Fig.  164  the  arc  F''-0'^ F%  with  center  at  B'^  and 

with  radius  equal  to  the  radius  of  the  sphere,  may  represent  the 
indefinite  development  of  the  curve  cut  from  the  surface  of  the 
cone  by  the  sphere. 

To  determine  the  positions  in  development  of  the  points  on  this 
curve  through  which  the  various  elements  of  the  cone  pass  before 
development,  we  must  determine  the  actual  distances  between 
these  points  measured  along  the  curve  before  development. 

To  do  this,  develop  the  horizontal  projecting  cylinder  of  the 
original  curve,  as  show  in  Fig.  163.    The  base  of  this  cylinder  is 

fr^r9.i 'ff'  ^^^  since  the  plane  of  the  base  is  perpendicular  to 

the  elements  of  the  cylinder,  the  curve  of  the  base  will  roll  out 
into  a  straight  line  to  which  the  elements  of  the  cylinder  will 
remain  perpendicular  (see  Problem  249,  Case  4). 

The  development  of  this  cylinder  is  shown  in  Fig.  165,  where 
F^-Fj  represents  the  development  of  the  curve  of  the  base,  and 

where  F-O-Q-  • F'  represents  the  development  of  the  original 

curve  of  intersection.  The  actual  distances  between  the  points  of 
division  on  this  curve  are  now  revealed. 

Returning  now  to  Fig.  164,  start  with  F'^\,  any  point  on  the  arc, 
and  make  the  distances  F''-0'\  0''-Q'\  Q'^-G'\  etc.,  such  that  their 
rectified  lengths  shall  be  equal  respectively  to  the  rectified  dis- 
tances F-0,  0-Q,  Q-G,  etc.,  of  Fig.  165.  Through  B''  and  the 
points  just  determined  draw  B''-F'^-D'\  B'^-(y^-N'\  etc.,  to  repre- 
sent in  development  the  elements  of  the  cone  passing  through  the 
points  F^  O,  Q,  etc.,  of  Fig.  163. 

Let  it  now  be  required  to  develop  the  curve  of  the  base  of  the 
cone.  In  Fig.  164  make  B''-iy\  B'-N\  B''~P'\  etc.,  equal  respec- 
tively to  B-D,  B-N,  B-P,  etc.,  of  the  original  cone,  Fig.  163.  The 
curve  D'^-N'^-P'^-B'^-D'j  represents  the  curve  of  the  base  of  the 
cone  in  development.  In  the  same  way  we  may  find  the  develop- 
ment of  any  curve  on  the  surface  of  the  cone. 

422.  Problem  291.  Find  the  intersection  of  a  sphere  hy  a  cone 
when  the  axis  of  the  cone  contains  the  center  of  the  sphere,  and  when 
the  vertex  of  the  cone  is  outside  the  surface  of  the  sphere. 


INTERSEC^TIO^^  OF  SURFACES  BY  SURFACES   191 


423.  Problem  292.  Find  the  inter sectioyi  of  a  sphere  mid  a  cone 
when  the  vertex  of  the  cone  lies  outside  the  surface  of  the  sphere,  and 
when  some  of  the  elements  of  the  cone  lie  wholly/  outside  the  sphere. 

424.  Problem  293.  Given  a  sphere  and  a  point  without  the  sur- 
face; required  to  determine  a  cone  tvith  vertex  at  the  point  and 
with  surface  tangent  to  the  sphere,  and  to  determine  the  circle  of 
tangency  between  the  two  surfaces. 

425.  Problem  294.  To  find  the  intersection  of  any  tivo  surfaces 
of  revolutio7i  when  their  axes  intersect,  and  to  draw  a  rectilinear 
tangent  to  the  cu7've  of 
intersection. 

Case  1.  To  find 
the  intersection. 

Analysis.  Draw  a 
series  of  auxiliary 
spheres  with  centers 
at  the  point  in  which 
the  axes  intersect. 

These  spheres  will 
cut  from  each  surface 
the  circumference  of 
a  circle  whose  plane 
will  be  perpendicular 
to  the  axis  of  the  sur- 
face. The  circumfer- 
ences of  these  circles 
will  intersect  in  points 
common  to  both  sur- 
faces, and  therefore  in 
their  curve  of  inter- 
section. 

Construction.  Let  it 
be  required  to  find  the 
intersection  of  an  inverted  right  circular  cone  by  a  right  circular 
cylinder,  as  represented  in  Fig.  166.  A-B  represents  the  axis  of 
the  cone  and  D-E  represents  the  axis  of  the  cylinder,  the  two 
axes  intersecting  at  D  and  lying  in  a  plane  parallel  to  V. 


Fig.  166 


192  DESCRIPTIVE  GEOMETRY 

An  auxiliary  sphere  with  center  at  B  and  with  radius  equal  to 
^^-f  will  cut  the  surface  of  the  cone  in  the  circumference  of  a 
circle  whose  vertical  projection  is  f-g^  and  whose  horizontal  pro- 
jection i^f-m-g-Uj.  The  same  sphere  will  cut  the  surface  of  the 
cylinder  in  the  circumference  of  another  circle  whose  vertical  pro- 
jection is  h'-V.  The  circumferences  of  these  two  circles  intersect  in 
two  points  vertically  projected  at  (m',  71')  and  horizontally  pro- 
jected at  nij  and  n^  respectively.  M  and  N  are  two  points  of  the 
required  curve  of  intersection,  and  others  may  be  found  in  the 
same  way. 

The  meridian  plane  of  the  axes  will  cut  the  cone  and  the  cylin- 
der in  those  elements  which,  when  projecting  upon  F,  appear  as 
extreme  elements.  These  elements  intersect  at  0  and  P  respec- 
tively, the  lowest  and  the  highest  points  of  the  curve  of  intersection. 

When  projecting  on  i/,  if  we  regard  the  inverted  cone  as  hollow, 
that  portion  of  the  curve  of  intersection  lying  on  the  upper  surface 
of  the  cylinder  will  be  visible. 

When  projecting  on  F,  that  portion  of  the  curve  of  intersection 
lying  on  the  front  surface  of  the  cylinder  and  on  the  front  surface 
of  the  cone  will  be  visible. 

Case  2.    To  draw  a  rectilinear  tangent  to  the  curve  of  intersection. 

The  rectilinear  tangent  to  the  curve  of  intersection  at  any  point 
will  be  the  intersection  of  two  planes,  one  tangent  to  the  cone 
at  the  given  point  and  the  other  tangent  to  the  cylinder  at  the 
same  point. 

By  use  of  methods  already  explained  we  may  develop  both  of 
the  given  surfaces  and  show  the  character  of  the  curve  of  inter- 
section when  rolled  out  into  a  plane  surface. 

426.  Problem  295.  By  the  process  just  explained  find  the  inter- 
section of  two  cylinders  of  revolution  whose  axes  intersect. 

427.  Problem  296.  By  the  above  process  find  the  intersection  of  a 
cone  of  revolution  and  a  sphere. 

428.  Problem  297.  By  the  above  process  find  the  intersection  of 
two  cylinders  of  revolution^  axes  intersecting^  the  larger  horizontal 
and  the  smaller  vertical. 


CHAPTER  XVIII 


ISOMETRIC  PROJECTION  AND   OTHER  FORMS   OF   ONE-PLANE 

PROJECTION 

429.  Introductory  Statements.  In  making  a  working  drawing  of 
a  solid  it  is  customary  to  place  the  object  in  such  a  position  that 
the  plane  of  two  of  its  dimension  lines  is  parallel  to  the  plane  of 
projection.  This  results  in  the  rep- 
resentation of  only  one  face  of  the 
object  at  a  time. 

It  is  sometimes  desirable,  how- 
ever, to  show  three  dimension  faces 
of  a  solid  in  one  view,  and  without 
going  to  the  trouble  of  making  a 
perspective  drawing.  This  is  ac- 
complished by  isometric  projection. 

In  Fig.  167  let  0-X,  O-F,  and 
0-Z  represent  the  three  edges  of 
a  right  trihedral  angle  with  vertex 
at  0.  Through  the  vertex  0  draw 
a  straight  line  P—O-o'  equally  in- 
clined to  the  three  edges.  Through  any  point  o'  on  P-O-o'  pass 
a  plane  F  perpendicular  to  P-0-o\  and  project  the  three  edges  of 
the  right  trihedral  angle  upon  this  plane.  Since  the  edges  them- 
selves are  equally  inclined  to  the  line  P-0-o\  and  therefore  equally 
inclined  to  the  plane  F,  and  since  they  form  the  three  edges  of  a 
right  trihedral  angle,  the  three  projections  o -a;',  o^-y\  and  o^-z^ 
will  radiate  from  o^  so  as  to  form  angles  of  120  degrees. 

The  lines  0-X^  0-Y,  and  0-Z  are  called  coordinate  axes.  The 
plane  Fis  called  the  isometric  plane  of  projection  ;  the  point  o\  the 
projection  of  0,  is  called  the  isometric  origin;  and  the  lines  o'-x', 
o'-y\  and  o'-z',  which  are  the  projections  upon  the  plane  F  of  the 
original  axes  0-X,  O-Y,  and  0-Z,  are  called  the  isometric  axes. 

193 


Fig.  167 


194  DESCRIPTIVE  GEOMETEY 

The  planes  determined  by  the  lines  0-X  and  0-F,  0-Y  and 
0-Z^  and  0-Z  and  0-X  are  called  coordinate  plafies^  and  the  pro- 
jections of  magnitudes  lying  on  these  planes  between  the  limiting 
axes  will  fall  between  the  projections  of  these  limiting  axes. 

The  axis  0-Y  is  given  a  position  directly  beneath  the  line  P-0. 
As  a  result  the  isometric  axis  o'-«/'  has  a  vertical  position,  the 
axis  o'-x'  makes  an  angle  of  30  degrees  with  a  horizontal  line 
on  the  right,  and  the  axis  o'-z'  makes  an  angle  of  30  degrees  with 
a  horizontal  line  on  the  left.  For  this  reason  the  isometric  axes 
and  all  straight  lines  parallel  to  them  are  easily  represented 
upon  the  drawing  board  by  use  of  the  T-square  and  the  30-degree 
triangle. 

Each  of  the  three  dimensions  —  length,  breadth,  and  thick- 
ness —  of  a  solid  are  measured  on  straight  lines  perpendicular  to 
the  plane  of  the  other  two,  corresponding  with  the  three  edges  of 
a  right  trihedral  angle. 

It  is  therefore  possible  to  place  a  solid  in  such  a  position  that 
its  three  principal  dimension  lines  shall  coincide  with  the  three 
coordinate  axes  0-X^  0-T,  and  0-Z  of  Fig.  167.  When  the  mag- 
nitude occupies  this  position  its  projection  upon  the  plane  V  will 
reveal  the  characteristics  of  three  dimension  faces  of  the  object, 
and  the  projection  is  called  isometric. 

Whenever  a  solid  is  projected  upon  a  plane  to  which  its  prin- 
cipal dimension  lines  are  equally  inclined,  the  projection  is  called 
isometric,  since  on  account  of  this  equality  of  inclination  the  pro- 
jections of  equal  distances  measured  upon  these  dimension  lines, 
or  upon  lines  parallel  to  them,  will  be  equal. 

The  projection  of  any  magnitude  may  be  said  to  be  isometric 
whenever  this  projection  is  determined  by  reference  to  isometric 
axes  which  themselves  are  the  isometric  projections  of  the  three 
coordinate  axes  to  which  the  magnitude  is  referred,  coordinates 
with  reference  to  the  isometric  axes  appearing  as  the  isometric 
projections  of  the  corresponding  coordinates  used  in  connection 
with  the  original  coordinate  axes. 

430.  The  Isometric  Scale.  The  inclination  of  each  of  the  three 
coordinate  axes  0-X,  0-Y^  and  0-Z  to  the  plane  of  projection  is 
35°. 16'.    By  trigonometry  the   isometric  projection   of   one  foot 


ISOMETRIC  PROJECTION  195 

measured  upon  either  one  of  these  axes  is  1  foot  times  the  natural 
cosine  of  35°. 16',  which  is  equal  to  .81647  ft. 

If  this  length  is  divided  into  twelve  equal  parts,  each  one  of 
these  divisions  will  represent  the  isometric  projection  of  an  inch 
measured  under  the  conditions  mentioned  above.  In  the  same 
way  we  may  subdivide  the  inch  and  thus  establish  an- isometric 
scale  by  which  we  may  determine  the  length  in  isometric  pro- 
jection of  any  distance  measured  upon  the  coordinate  axes  or  upon 
any  lines  parallel  to  these  axes. 

Since  drawings  are  usually  made  either  on  a  smaller  or  a  larger 
scale  than  tlie  object  represented,  and  since  on  account  of  the  equal 
inclination  of  the  coordinate  axes  to  the  plane  of  projection,  the  pro- 
jections of  distances  measured  along  these  lines  are  equally  fore- 
shortened, there  is  no  good  reason  why  any  other  than  the  ordinary 
foot  scale  should  be  used  in  connection  with  this  form  of  projection. 
The  isometric  scale  is  therefore  not  used  in  practical  drafting. 

431.  Shade  Lines.  In  isometric  projection  the  rays  of  light  are 
assumed  parallel  to  the  plane  •  of  projection  and  inclined  at  an 
angle  of  30  degrees  with  the  horizon.  With  these  exceptions, 
shade  lines  in  isometric  projection  are  determined  and  represented 
precisely  as  in  orthographic  projection. 

If  the  cube  to  whose  diagonal  rays  of  light  in  orthographic  pro- 
jection have  been  referred  (see  Section  230)  is  so  placed  that  the 
diagonal  passing  through  the  upper  right-hand  front  vertex  of  the 
cube  is  perpendicular  to  F,  and  so  that  the  right-hand  vertical  edge 
of  the  cube  is  still  in  a  vertical  plane,  the  diagonal  passing  through 
the  upper  left-hand  front  vertex  of  the  cube  in  its  first  position  — 
the  diagonal  to  which  rays  of  light  in  orthographic  projection 
were  referred  —  will  in  the  new  position  be  parallel  to  V  and  make 
an  angle  of  30  degrees  with  H, 

In  isometric  projection,  then,  rays  of  light  may  be  referred  to  the 
same  diagonal  of  the  cube  as  in  orthographic  projection,  provided 
the  cube  occupies  the  position  indicated  above. 

432.  General  Instructions  in  regard  to  Solution  of  Problems. 
The  isometric  projection  of  a  point  whose  coordinates  with  ref- 
erence to  two  coordinate  axes  are  known  is  the  intersection  of  the 
isometric  projections  of  the  two  coordinate  lines  of  the  point. 


196 


DESCKIPTIVE   GEOMETEY 


The  isometric  projection  of  a  point  whose  coordinates  with  refer- 
ence to  three  coordinate  planes  are  known  is  the  intersection  of  the 
isometric  projections  of  the  three  coordinate  lines  of  the  point. 

The  isometric  projection  of  a  straight  line  is  the  straight  line  de- 
termined by  the  isometric  projections  of  two  points  of  the  given  line. 
Since  the  orthographic  projections  of  parallel  lines  are  parallel, 
the  isometric  projections  of  parallel  lines  will  be  parallel  whether 
the  lines  themselves  are  parallel  to  the  coordinate  axes  or  not. 

The  isometric  projection  of  a  curved  line  is  the  line  determined 
by  the  isometric  projections  of  the  points  which  determine  the  line. 

The  isometric  projection  of  a  sur- 
face determined  or  limited  by  lines, 
straight  or  curved,  is  that  surface  de- 
termined by  the  isometric  projections 
of  the  determining  or  limiting  lines. 

To  draw  the  isometric  projection 
of  a  magnitude  of  three  dimensions, 
place  the  magnitude  in  such  a  posi- 
tion that  its  three  principal  dimension 
lines  shall  be  either  coincident  with 
or  parallel  to  the  three  coordinate  axes 
0-X,  0-  F,  and  0-Z ;  then  the  isome1> 
ric  projections  of  the  principal  dimension  lines  of  the  magnitude 
will  be  either  coincident  with  or  parallel  to  the  isometric  axes. 

433.  Problem  298.  To  find  the  isometric  projection  of  a  point  whose 
coordinates  with  reference  to  two  rectangular  axes  are  known. 

In  Fig.  168  let  o'-x\  o'-y\  and  o'-z'  represent  the  isometric  axes. 
Place  the  two  rectangular  axes  to  which  the  point  is  referred 
in  coincidence  with  0-X  and  0-Z  respectively.    Then  o'-x'  and 
0 -z'  are  the  isometric  projections  of  the  rectangular  axes. 

Make  o'-a^  equal  to  the  distance  of  the  point  from  the  axis  0-Z, 
and  through  a'  draw  a'-d'  parallel  to  o'-z'. 

Make  o'-h'  equal  to  the  distance  of  the  point  from  the  axis 
0-X,  and  through  h'  draw  h'-d'  parallel  to  o'-x'. 

The  lines  a'-d'  and  h'-d'  are .  the  isometric  projections  of  the 
coordinate  lines  of  the  point,  and  the  point  d'  is  the  isometric 
projection  sought. 


y' 


Fig.  168 


ISOMETRIC  PROJECTION 


197 


Of  course  the  same  result  will  be  obtained  if  after  drawing  the  line 
a'-d'  we  lay  off  upon  it  the  remaining  coordinate  a'-d'  equal  to  o'-h'. 
The  axes  to  which  the  point  is  referred  may  be  assumed  in  coinci- 
dence with  0-X  and  0-  F,  or  in  coincidence  with  0-Z  and  0-  Y. 

434.  Problem  299.  To  find  the 
isometric  projection  of  a  point  whose 
coordinates  with  reference  to  three 
rectangular  coordinate  planes  are 
known. 

In  Fig.  169  let  o'-x\  o'-z/',  and  o'-z' 
represent  the  isometric  axes. 

Place  the  three  coordinate  axes, 
to  whose  planes  the  point  is  referred, 
in  coincidence  with  0-X,  0-F,  and 
0-Z  respectively.  Then  o'-x'^  o'-i/\ 
and  o'-z'  are  the  isometric  projections  of  the  coordinate  axes. 

Make  o'-a'  equal  to  the  distance  of  the  point  from  the  plane 
^-0-  rand  through  a'  draw  a'-b'  parallel  to  o'-z'.  Make  a'-b'  equal 
to  the  distance  of  the  point  from  the  plane  X-O-Y,  and  through 
b'  draw  b'-d^  parallel  to  o'-y'.    Make  b'-d'  equal  to  the  distance  of 

the  point  from  the  plane  Z-O-X, 
The  point  d'  is  the  isometric  pro- 
jection of  the  point  in  question. 

The  lines  e'-d',  d'-f,  and  b'-d' 
are  the  isometric  projections  of 
the  three  coordinate  lines  of  the 
point  D. 

435.  Problem  300.  To  fijid  the 
isometric  projection  of  a  straight 
line  when  the  coordiiiates  of  two  of 
its  points  with  reference  to  three 
rectangular  coordinate  planes  are 
known. 

In  Fig.  170  let  o'-x',  o'-y\  and  o'-z'  represent  the  isometric  axes. 
Place  the  coordinate  axes  to  whose  planes  the  line  is  referred 
in  coincidence  with  0-X,  0-  F,  and  0-Z.    Then  o'-x^,  o'-y',  and  o'-z' 
are  the  isometric  projections  of  these  coordinate  axes. 


Fig.  170 


198 


DESCPaPTIVE   GEOMETRY 


Fig.  171 


By  Section  43-i  find  m'  and  n\  the  isometric  projections  of  the 
two  given  points.    The  line  m'-n'  is  the  required  projection. 

436.  Problem  301.  To  faid  the  iso metric  jjrojection  of  a  square 
card  when  the  coordinates  of  its  four  vertices  with  reference  to  two 

rectangidar  axes  are  knoiun. 

In  Fig.  171  let  o'-j'  and  o'-z' 
represent  the  isometric  projec- 
tions of  the  two  axes  to  which 
the  card  is  referred. 

By  Section  433  find  the  iso- 
metric projections  of  the  four 
vertices  Ji,  JV,  P,  and  Q. 

The  polygon  m'-n'-p'-q'  is  the 
projection  sought. 
The  projection  may  be  made  with  reference  to  the  axes  o'-x' 
and  o'-y\  or  with  reference  to  the  axes  o'-y'  and  o'-z'. 

437.  Problem  302.  To  find  the  isometric  projection  of  a  triangular 
card  when  the  coordinates  of  its  three  vertices  with  reference  to  three 
rectangular  coordinate  planes  are  known. 

In  Fig.  172  let  o'-x\  o'-y'.,  and  o'-z'  represent  the  isometric 
projections  of  the  axes  to  whose  planes  the  card  is  referred. 

By  Section  434  find  the  iso- 
metric projections  of  the  three 
vertices  Jf,  iV,  and  F.  The  tri- 
angle m'—n'—p'  is  the  projection 
sought. 

438.  Problem  303.  To  make  an 
isometric  projection  of  a  cube. 
I  See  Fig.  173.  Let  o'-x',  o'-y', 
and  o'-z'  represent  the  isometric 
axes.  Place  the  cube  so  that  one 
of  its  vertices  shall  coincide  with 
the  origin  of  coordinates,  0,  and  so  that  its  three  adjacent  edges 
shall  coincide  with  the  three  coordinate  axes  0-X,  0-Y,  and  0-Z. 

The  isometric  projections  of  these  three  edges  will  fall  on  the 
isometric  axes,  and  the  isometric  projections  of  the  remaining 
edges  will  be  parallel  to  these  three  axes. 


Fig.  172 


ISOMETRIC   PROJECTION 


190 


Make  o'-a'  and  o'-d!  each  equal  to  the  edge  of  the  cube. 
Through  a'  draw  a'-e'  parallel  to  o'-d'.  Through  d'  draw  d'-e' 
parallel  to  o'-a'  and  intersecting  a'-e'  at  e'.  The  figure  o'-a'-e'-d' 
represents  the  isometric  projection  of  the  upper  face  of  the  cube. 

Make  o'-h'  equal  to  the  edge  of  the  cube.  Through  a'  draw 
a'-f  parallel  to  o'-h'.  Through 
h'  draw  h'-f  parallel  to  o'-a'  and 
intersecting  a'-f  at  /'.  The  fig- 
ure o'-a'-f'-b'  represents  the  iso- 
metric projection  of  the  right-hand 
face  of  the  cube. 

Through  h'  draw  h'-g'  parallel 
to  o'-d'.  Through  d'  draw  d'-g' 
parallel  to  o'-h'  and  intersecting 
h'-g'  at  g'.  The  figure  o'-h'-g'-d' 
represents  the  isometric  projection 
of  the  left-hand  face  of  the  cube. 

The  figure  e' -a' -f -h' -g' -d'  is  the  isometric  projection  of  the 
cube. 

To  determine  the  shade  lines  of  the  cube. 

With  the  source  of  light  in  its  new  position,  as  stated  above, 

the  three  faces  0-E^  0-G,  and 
G-E  will  be  in  the  light,  and  the 
three  remaining  faces  will  be  in 
the  dark.  Therefore  the  visible 
shade  lines  are  G-B^  B-0^  0-A, 
and  A-E. 

439.  Problem  304.  To  make  an 
isometric  projection  of  a  square 
pedestal  supporting  a  square  pris- 
matic pillar. 

See  Fig.  174.  Place  the  upper 
front  vertex  of  the  pedestal  at  the 
origin  0.  Make  o'-a^  and  o'-h'  each  equal  to  the  side  of  the  square 
pedestal ;  also  make  o'-d'  equal  to  the  altitude  of  the  pedestal. 
The  pedestal  is  now  completed  by  drawing  through  a',  b',  and  d' 
straight  lines  parallel  to  the  proper  axes. 


200 


DESCKIPTIVE  GEOMETKY 


Knowing  the  relation  of  the  horizontal  dimensions  of  the  pillar 
to  the  horizontal  dimensions  of  the  pedestal,  we  can  determine  the 
distance  between  the  corresponding  faces  of  the  pillar  and  pedestal. 
Make  o'-e'  equal  to  the  distance  between  the  left-hand  face  of  the 
pillar  and  the  left-hand  face  of  the  pedestal,  and  through  e'  draw 
e'-f-k'  parallel  to  o'-z'.  Make  e'-f  equal  to  the  distance  between 
the  right-hand  face  of  the  pillar  and  the  right-hand  face  of  the 
pedestal,  and  through  /'  draw  f'-g'  parallel  to  o'-x' .    Make  f'-k' 

ajidf'-g'  each  equal  to 


the  side  of  the  square 
pillar.  Through  /', 
k\  and  g^  draw  the 
vertical  straight  lines 
/-^^Ar'-?^^  and/-77^^ 
Make  f-V  equal  to 
tlie  altitude  of  the 
pillar  and  complete 

the  drawing. 

440.  Problem  305.    To    make    an 

isometric  projection  of  an  hexagonal 

prism. 

Analysis.    Imagine  the  hexagonal 


Fig.  175 


prism  to  be  inscribed  within  a  rec- 
tangular prism  whose  rectangular  bases  circumscribe  the  hexag- 
onal bases  of  the  given  prism,  and  whose  altitude  is  equal  to 
that  of  the  given  prism. 

Draw  the  isometric  projection  of  the  rectangular  prism  and  then 
by  reference  to  this  prism  make  the  isometric  projection  of  the 
hexagonal  prism. 

Construction.    See  Figs.  175  and  176.    Let  L-M-N-P i, 

Fig.  176,  represent  the  hexagonal  base  of  the  given  prism,  and 
let  0-D-E-A^  of  the  same  figure,  represent  the  circumscribing 
rectangle  of  this  base. 

The  isometric  projection  of  the  circumscribing  prism  is  repre- 
sented in  Fig.  175  by  o' -a^ -e^ -d' -h\  where  o'-a'  and  o'-d'  are  equal 
respectively  to  the  horizontal  dimensions  of  this  prism,  and  where 
o'-h'  is  equal  to  the  altitude  of  the  prism. 


ISOMETRIC  PROJECTION 


201 


To  draw  the  isometric  projection  of  the  upper  base  of  the  hex- 
agonal prism,  make  o'-Z',  o'—m\  d'-n\  etc.,  of  Fig.  175  equal  respec- 
tively to  0-X,  0-M^  D-N^  etc.,  of  Fig.  176,  and  connect  the  points 
thus  found  by  straight  lines. 

To  complete  the  projection  of  the  prism,  find  the  isometric 
projection  of  the  lower  base  by  the  process  just  explained  and 
connect  the  corresponding  vertices  of  the  two  bases. 

The  shade  lines. 

The  upper  base  and  the  three  faces  L-R^  M-  f/,  and  TV-  W  are  in 
the  light,  locating  the  shade  lines  as  shown  in  the  drawing. 


Fig.  177 


441.  Problem  306.   To  make 

an  isometric  projection  of  the 

frustum    of  an   octagonal 

pyramid. 

Analysis.    Imagine  the  frustum  of  the  pyramid  to  be  inscribed 

within  a  square  prism  whose  base  is  the  circumscribing  square  of 

the  larger  base  of  the  frustum  and  whose  altitude  is  the  altitude 

of  the  frustum. 

Construction.  See  Figs.  177  and  178.  Fig.  178  represents  in 
plan  the  two  bases  of  the  frustum  and  their  circumscribing  squares. 
B-G-K-F  represents  the  circumscribing  square  of  the  larger  base 

and  L-M-JSf-P represents  the  larger  base  itself. 

Q-E-U-JV  represents  the  circumscribing  square  of  the  smaller 
base  and  1-2-3-4 represents  the  smaller  base  itself. 


202 


DESCKIPTIVE   GEOMETKY 


The  isometric  projection  of  the  circumscribing  prism  is  repre- 
sented in  Fig.  177  by  o' -a' -e' -d' -h' ,  where  o'-a'  and  o'-d'  are  each 
made  equal  to  the  side  of  the  circumscribing  square  of  the  larger 
base  of  the  frustum,  and  where  o'-U  is  made  equal  to  the  altitude 
of  the  frustum. 

The  isometric  projection  of  the  lower  base  of  the  frustum  is  repre- 
sented in  Fig.  177  by  V -m' -'nJ -p' -  •  •  •,  where  h'-V^  h'-7n\  b'-n'-,  etc., 
are  made  equal  respectively  to  i?-Z,  B-M,  B-N^  etc.,  of  Fig.  178. 

The  isometric  projection  of  the  circumscribing  square  of  the 
upper  base  of  the  frustum  is  represented  in  Fig.  177  by  q'-r'-ii'-w\ 
concentric  with  o'-d'-e'-a'  and  having  its  side  equal  to  the  side  of 
the  square  Q-B-U-W  oi  Fig.  178. 

The  isometric  projection  of  the  upper  base  of  the  frustum  is 
represented  in  Fig.  177  by  1  — 2'-3'-4'— •  •  •,  determined  as  previ- 
ously explained. 

The  isometric  projection  of  the  edges  of  the  frustum  may  now 
be  represented  by  connecting  the  vertices  of  the  upper  base  with 
the  corresponding  vertices  of  the  lower  base. 

442.  Problem   307.     To 
make  an  isometric  drawing 


Fig.  180 
of  a  box  with  open  lid. 

Construction.    See    Figs. 
179  and  180.    The  isomet- 
ric projection    of   the   box 
is  represented  in  Fig.  179  by  o'-a'-e'-d'-b'. 

Fig.  180  represents  an  end  view  of  the  box  and  lid.    By  reference 
to  a  horizontal  and  a  vertical  axis  through  D  in  this  figure  we  can 


ISOMETEIC  PROJECTION 


203 


determine  the  coordinates  of  any  point  on  the  lid  with  reference  to 
a  horizontal  and  to  a  vertical  plane  through  the  hinge  axis  D-E. 

The  point  F^  for  example,  one  apex  of  the  lid,  is  at  the  distance 
F-g  above  the  horizontal  plane  and  at  the  distance  F-k  back  of 
the  vertical  plane.  The  point  F  is  also  in  the  plane  of  the  left-hand 
end  of  the  box,  and  in  isometric  projection  will  be  located  with 
reference  to  the  isometric  axes  o'-z'  and  o'-y' . 

In  Fig.  179  make  d'-g'  equal  to  D-g  of  Fig.  180,  and  through 
g'  draw  g'-f  parallel  to  the  axis  o'-y'.  Lay  off  upon  g'-f  from  g' 
the  distance  g-F  of  Fig. 
180.  The  point  /  is  the 
isometric  projection  of  F. 

In  this  way  the  isomet- 
ric projection  of  any  point 
on  the  lid  may  be  deter- 
mined. 

The  isometric  projec- 
tion of  aS^,  one  of  the  inner 
vertices  of  the  lid,  is  found 
thus :  In  the  end  view, 
Fig.  180,  the  point  S  will 
appear  at  P,  at  the  dis- 
tance P-q  below  the  hori- 
zontal plane  and  at  the 
distance  P-r  back  of  the 
vertical  plane. 

In  Fig.  179  make  d'-q'  and  q'-p'  equal  respectively  to  D-q  and 
q-P  of  Fig.  180,  locating  p' .  Through  p\  which  is  the  isometric 
projection  of  P,  draw  p'-s'  parallel  to  o'-x'^  to  represent  the  inner 
horizontal  edge  of  the  lid. 

Through  t'  previously  located  draw  t'-s'  parallel  to  f'-l'^  to 
represent  the  inner  vertical  edge  of  the  lid. 

The  point  s',  the  intersection  of  p'-s'  and  t'-s\  is  the  isometric 
projection  of  S. 

443.  Problem  308.    To  represent  a  mortise  and  tenon.    See  Fig.  181. 

444.  The  Isometric  Projection  of  Curved  Lines.  The  isometric 
projection  of  curved  lines    may  be    determined    by  finding   the 


Fig.  181 


204 


DESCRIPTIVE  GEOMETRY 


isometric  projections  of  a  number  of  the  points  of  the  curve  and 
by  tracing  the  required  curve  through  these  projections. 

If  the  curve  is  of  single  curvature,  locate,  by  reference  to  two 
rectangular  axes,  a  sufficient  number  of  points  on  the  curve  to 
determine  the  curve  accurately.  Find  the  isometric  projections  of 
the  rectangular  axes  and  then  by  use  of  the  known  coordinate 
distances  determine  the  isometric  projections  of  the  points  which 
locate  the  curve. 

If  the  curve  is  of  double  curvature,  locate,  by  reference  to  three 
rectangular  coordinate  planes,  a  sufficient  number  of  points  on  the 
^^  D____G C 


Fig.  182 


curve  to  determine  it. 

Find  the  isometric  pro- 
jections of  the  coordinate 
axes  and  then  by  use  of 
the  known  coordinates, 
referred  to  these  axes, 
determine  the  isometric 
projections  of  the  points  which  locate  the  curve. 

445.  Problem  309.  To  make  an  isometric  drawing  of  a  circle. 
See  Figs.  182  and  183.  Fig.  183  represents  the  circle  referred 
to  the  four  sides  of  a  circumscribing  square  A-B-C-D,  any  two 
of  whose  adjacent  sides  may  be  taken  as  rectangular  axes  of  refer- 
ence. The  coordinates  of  a  number  of  points,  E,  Z,  F,  M,  G,  iV, 
JT,  P,  on  the  circumference,  including  the  points  of  tangency,  are 
here  determined. 

Fig.  182  shows  the  isometric  projection  of  the  circle  in  three 
positions.  First,  when  the  isometric  projections  of  the  axes  A—B  and 
A-D  coincide  with  the  isometric  axes  o'-x'  and  o'-z'  respectively. 
Second,  when  the  isometric  projections  of  the  axes  D-C  and  D-A 


ISOMETRIC  PROJECTION 


205 


coincide  with  the  isometric  axes  o'-x'  and  o'-y'  respectively.  Third, 
when  the  isometric  projections  of  the  axes  C-D  and  C-B  coincide 
with  the  isometric  axes  o'-z'  and  o'-y'  respectively. 

In  each  case  the  isometric  projections  of  the  points  are  found 
by  use  of  the  coordinate  measurements  made  in  Fig.  183.  For 
example,  the  point  p\  Fig.  182,  is  obtained  by  making  o'-w'  and 
w'-p'  equal  respectively  to  ^-/^Fand  W-P  of  Fig.  183. 


Fig.  185 


The  point  V"  is  located  by  making  o'-u'  and  tt'-Z'"  equal  respec- 
tively to  C-U  and  U-L  of  Fig.  183. 

446.  Problem  310.  To  make  an  isometric  draiving  of  any  plane 
curve. 

See  Figs.  184  and  185.  Let  Fig.  185  represent  the  curve  when 
referred  to  two  rectangular  axes,  one  of  which  is  0-A  tangent  to 
the  curve  at  D  and  the  other  0-B  tangent  to  the  curve  at  F, 

The  coordinates  of  a  number  of  points,  H",  R^  Z>,  F^  M,  Q,  on 
the  curve  are  determined  with  reference  to  these  axes. 

The  isometric  projection  of  this  curve  is  shown  in  Fig.  184, 
where  the  isometric  projections  of  the  axes  0-A  and  0-B  coincide 
with  the  isometric  axes  o'-y'  and  o'-x'  respectively,  and  where  the 
isometric  projections  of  the  points  W,  B,  Z>,  etc.,  are  found  as 
previously  explained. 

We  may  draw  the  isometric  projection  of  this  curve  when  the 
curve  occupies  any  desired  position  with  reference  to  the  coordinate 
axes  0-X,  0-  Y,  and  0-Z,  whether  in  the  plane  determined  by  any  two 
of  these  axes  or  in  the  space-angle  determined  by  the  three  axes. 


206 


DESCRIPTIVE  GEOMETRY 


447.  Problem  311.  To  make  an  isometric  drawing  of  any  curve  of 
double  curvature  when  referred  to  three  rectangular  coordinate  planes. 

Let  it  be  required  to  make  an  isometric  drawing  of  the  helix, 
when  the  helix  is  referred  to  the  three  orthographic  planes  of  pro- 
jection P,  F;  and  //. 

Fig.  187  represents  the  orthographic  projection  of  the  helix, 
from  which  the  three  coordinate  lines  of  any  point  of  the  curve 
may  be  determined. 

The  isometric  projection  of  the  curve  is  shown  in  Fig.  186,  where 
x'-o'-y'  represents  the  isometric  projection  of  the  plane  F,  where 


,^!\ 


Fig.  186 


x'-o'-z'  represents  the  isometric  projection  of  the  plane  H,  and 
where  z^-o'-y'  represents  the  isometric  projection  of  the  plane  P. 

The  isometric  axis  o'-x'  represents  the  isometric  projection  of 
G-L^  o'-z'  represents  the  isometric  projection  of  P-pp  and  o'-y' 
represents  the  isometric  projection  of  P-p'. 

The  isometric  projection  of  any  point  of  the  curve,  for  example 
the  point  Jf,  may  be  found  thus : 


OBLIQUE  PROJECTION  207 


Fig.  187  shows  that  the  distance  of  iTf  from  the  plane  Pis  w,- 


u 


that  its  distance  from  V  is  m,-w^  and  that  its  distance  from  If  is 


w-m'. 


In  Fig.  186  make  o'-w'  equal  to  the  distance  of  Jf  from  the  plane 
P,  and  through  w'  draw  w'-r'  parallel  to  o'-z'.  Make  w'-r'  equal 
to  the  distance  of  M  from  F,  and  through  r'  draw  r^-m'  parallel  to 
o'-i/'.  Make  r'-m'  equal  to  the  distance  of  M  from  H.  The  point 
m'  is  the  isometric  projection  of  Jf,  and  other  points  may  be  found 
in  the  same  way. 

448.  Oblique  Projection.  In  orthographic  and  in  isometric  projec- 
tion the  observer  is  assumed  at  an  infinite  distance  from  the  plane 
of  projection,  and  the  resultant  parallel  projecting  lines  are  taken 
perpendicular  to  the  plane  of  projection. 

If  the  projecting  lines  are  taken  oblique  to  the  plane  of  projec- 
tion, the  system  of  projection  is  called  oblique  projection, 

449.  Cavalier  Perspective  or  Cabinet  Projection.  When  the  project- 
ing lines  are  assumed  parallel  to  each  other  and  at  an  inclination  of 
45  degrees  to  the  plane  of  projection,  the  system  of  projection  is 
called  cavalier  perspective  or  cabinet  projection. 

In  Fig.  188  let  a'-B  represent  a  straight 
line  perpendicular  to  V  and  piercing  it  at  a'. 
Through  B  draw  a  projecting  line  B-b'  at  an  in- 
clination of  45  degrees  to  Tand  piercing  Fat  5'. 
The  line  a'-b'  is  the  cabinet  projection  of  a'-B, 

No  restriction  whatever  is  made  upon  the 
direction  which  the  projecting  lines  shall  take 
so  long  as  they  are  inclined  at  an  angle  of  45  degrees  with  the  plane 
of  projection.  In  any  given  problem,  however,  the  projecting  lines 
must  be  parallel. 

450.  Observations.  Since  the  projecting  lines  are  inclined  45 
degrees  to  the  plane  of  projection,  the  cabinet  projection  of  a 
straight  line  perpendicular  to  the  plane  of  projection  will  be  equal 
to  the  line  itself. 

The  cabinet  projection  of  a  straight  line  in  the  plane  of  projec- 
tion will  be  the  line  itself. 

The  cabinet  projection  of  a  straight  line  parallel  to  the  plane  of 
projection  will  be  equal  to  the  line  itself. 


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